\hr SSKJr SSKJr SSKJr SSKJr SSKJ r SSK J r SSK J r SSKJrJr SS KJrJrJrJrJr SS KJr SS KJr SS KJrJr SS KJrJ r SSK!J"r" SSK#J$r$J%r%J&r&J'r' SSK(J)r) SSK*J+r+ SSK,J-r- SSK.J/r0 SSK1J2r2J3r3J4r4 SSK5J6r6 SSK7J8r8J9r9 SSK:J;r; SSKr> SSK?J@r@ SSKAJBrBJCrC SSKDJErE SSKFJGrG SSKHJIrI SSKJJKrKJLrL SSKMJNrNJOrOJPrPJQrQJRrR S S!/rS"S"S#\T5rU"S$S%5rV"S&S'\V5rW"S(S)\V5rX"S*S+\V5rY"S,S-\V5rZ"S.S/\V5r["S0S1\V5r\"S2S3\V5r]"S4S5\V5r^"S6S7\V5r_"S8S9\V5r`"S:S;\V5ra"S<S=\V5rb\X\W\Y\Z\[\\\]\^\_\`\a\b/ rc\cVs1sHoRiM snreS>rfS?rgS@rhSAriSBrj\ "SCSDSE9SFSG4SHjrkSIrlS|SJjrm\ "SCSDSE94SKjrnS}SLjroSMSNR\q"\e55-\olr\ "SCSDSE94SOjrsS|SPjrtSQruSRrv\ "SCSDSE94SSjrwSTrx\ "SCSDSE94SUjrySVrzS~SXjr{SYr|\ "SCSDSE94SZjr}S[r~S\rS]rS^rS_rS`rSarSSbjrScrSdrSerSfrSgrSShjrSirSSjjrSkrSlrSmrSnrSSojrSpr\ "SWSDSE94SqjrSSrjrSSsjrSStjrSurSvrSwrSSxjr\rSyrSSzjrS{rgFs snf)) annotations)Add)check_assumptions)Tuple) factor_terms)_mexpand)Mul)Rational int_valued)igcdexilcmigcdinteger_nthrootisqrt)Eq)S)default_sort_keyordered)Symbolsymbols)_sympify)jacobiremoveinvertiroot)sign)floor)sqrt)MutableDenseMatrix)divisors factorint perfect_power) nextprime) is_squareisprime)symmetric_residue)sqrt_mod sqrt_mod_iter)GeneratorsNeeded)Poly factor_list)signsimp) solveset_real)numbered_symbols)as_int filldedent) is_sequencesubsets permute_signssigned_permutationsordered_partitions diophantine classify_diopcP^\rSrSrSrU4SjrU4SjrSrSrSr Sr S r U=r $) DiophantineSolutionSet(a@ Container for a set of solutions to a particular diophantine equation. The base representation is a set of tuples representing each of the solutions. Parameters ========== symbols : list List of free symbols in the original equation. parameters: list List of parameters to be used in the solution. Examples ======== Adding solutions: >>> from sympy.solvers.diophantine.diophantine import DiophantineSolutionSet >>> from sympy.abc import x, y, t, u >>> s1 = DiophantineSolutionSet([x, y], [t, u]) >>> s1 set() >>> s1.add((2, 3)) >>> s1.add((-1, u)) >>> s1 {(-1, u), (2, 3)} >>> s2 = DiophantineSolutionSet([x, y], [t, u]) >>> s2.add((3, 4)) >>> s1.update(*s2) >>> s1 {(-1, u), (2, 3), (3, 4)} Conversion of solutions into dicts: >>> list(s1.dict_iterator()) [{x: -1, y: u}, {x: 2, y: 3}, {x: 3, y: 4}] Substituting values: >>> s3 = DiophantineSolutionSet([x, y], [t, u]) >>> s3.add((t**2, t + u)) >>> s3 {(t**2, t + u)} >>> s3.subs({t: 2, u: 3}) {(4, 5)} >>> s3.subs(t, -1) {(1, u - 1)} >>> s3.subs(t, 3) {(9, u + 3)} Evaluation at specific values. Positional arguments are given in the same order as the parameters: >>> s3(-2, 3) {(4, 1)} >>> s3(5) {(25, u + 5)} >>> s3(None, 2) {(t**2, t + 2)} c>[TU]5 [U5(d [S5e[U5(d [S5e[ U5Ul[ U5Ulg)Nz$Symbols must be given as a sequence.z'Parameters must be given as a sequence.)super__init__r1 ValueErrortupler parameters)self symbols_seqr@ __class__s ]/var/www/auris/envauris/lib/python3.13/site-packages/sympy/solvers/diophantine/diophantine.pyr=DiophantineSolutionSet.__init__fsT ;''CD D:&&FG G[)  +c>[U5[UR5:wa/[S[UR5<S[U5<35e[U5n[ [U55HZnXn[ U5[ LdMU*R(dM/U*U;dM8UVs/sHoURU*U5PM nnM\ [TU])[U65 gs snf)Nz!Solution should have a length of z, not ) lenrr>setrangetypeint is_Symbolsubsr<addr)rAsolutionargsix_rCs rDrODiophantineSolutionSet.addrs x=C - -cRVR^R^N_ademanop p8}s8}%A A7c>rnnn!43;<8aFFA2qM8<&  E8$%=s.C&c8UHnURU5 M gN)rO)rA solutionsrPs rDupdateDiophantineSolutionSet.update}s!H HHX "rFc#r# [U5H$n[[URU55v M& g7frW)rdictzipr)rArPs rD dict_iterator$DiophantineSolutionSet.dict_iterators) Hs4<<23 3&s57c[URUR5nUH$nURUR"U0UD65 M& U$rW)r9rr@rOrN)rArQkwargsresultrPs rDrNDiophantineSolutionSet.subss?' dooFH JJx}}d5f5 6 rFc0[U5[UR5:a/[S[UR5<S[U5<35e[URU5VVs0sH up#UcM X#_M nnnUR U5$s snnf)NzEvaluation should have at most z values, not )rHr@r>r]rN)rArQpvreps rD__call__DiophantineSolutionSet.__call__sx t9s4??+ +SVW[WfWfSgilmqirst t #DOOT :L :atqt :Lyy~Ms + B8B)r@r) __name__ __module__ __qualname____firstlineno____doc__r=rOrYr^rNrh__static_attributes__ __classcell__)rCs@rDr9r9(s,;z , &4 rFr9cl\rSrSr%SrS\S'S SjrSr\S5r \S 5r SSS jjr S S jr S r g)DiophantineEquationTypea Internal representation of a particular diophantine equation type. Parameters ========== equation : The diophantine equation that is being solved. free_symbols : list (optional) The symbols being solved for. Attributes ========== total_degree : The maximum of the degrees of all terms in the equation homogeneous : Does the equation contain a term of degree 0 homogeneous_order : Does the equation contain any coefficient that is in the symbols being solved for dimension : The number of symbols being solved for strnameNc[U5RSS9UlUbX lOA[ URR5UlURR [ S9 UR(d [S5eURR5Ul [SURR555(d [S5e[UR5R5UlSUR;Ul[!UR5[!UR5-(+Ul[%UR5UlSUlg)NTforcekeyz+equation should have 1 or more free symbolsc38# UHn[U5v M g7frWr ).0cs rD 3DiophantineEquationType.__init__..s>*=Q:a==*=zCoefficients should be Integers)rexpandequation free_symbolslistsortrr>as_coefficients_dictcoeffallvalues TypeErrorr* total_degree homogeneousrIhomogeneous_orderrH dimension _parameters)rArrs rDr= DiophantineEquationType.__init__s *111=  # ,  $T]]%?%? @D     " "'7 " 8  JK K]]779 >$***;*;*=>>>=> > /<<>DJJ.&)$**oDrurHrr)rAr@s rD pre_solve!DiophantineEquationType.pre_solves`||~~QTXT]T]]^ ^  !:$"3"33 $J[J[]`ak]l!mnn%rF)rrrrrrrrrWNNreturnr9)rjrkrlrmrn__annotations__r=rpropertyrr@rrrorrFrDrrrrsK. I ,   T&rFrrc,\rSrSrSrSrSrSSjrSrg) Univariatea Representation of a univariate diophantine equation. A univariate diophantine equation is an equation of the form `a_{0} + a_{1}x + a_{2}x^2 + .. + a_{n}x^n = 0` where `a_{1}, a_{2}, ..a_{n}` are integer constants and `x` is an integer variable. Examples ======== >>> from sympy.solvers.diophantine.diophantine import Univariate >>> from sympy.abc import x >>> Univariate((x - 2)*(x - 3)**2).solve() # solves equation (x - 2)*(x - 3)**2 == 0 {(2,), (3,)} univariatec URS:H$Nrrrs rDrUnivariate.matchess~~""rFNcURU5 [URURS9n[ UR URS5R [R5HnURU45 M U$)Nr@r) rr9rr@r-r intersectrIntegersrO)rAr@rrbrRs rDrUnivariate.solvesg z"'(9(9dooVt}}d.?.?.BCMMajjYA JJt Z rFrr rjrkrlrmrnrurrrorrFrDrrs" D#rFrc,\rSrSrSrSrSrSSjrSrg) Lineara Representation of a linear diophantine equation. A linear diophantine equation is an equation of the form `a_{1}x_{1} + a_{2}x_{2} + .. + a_{n}x_{n} = 0` where `a_{1}, a_{2}, ..a_{n}` are integer constants and `x_{1}, x_{2}, ..x_{n}` are integer variables. Examples ======== >>> from sympy.solvers.diophantine.diophantine import Linear >>> from sympy.abc import x, y, z >>> l1 = Linear(2*x - 3*y - 5) >>> l1.matches() # is this equation linear True >>> l1.solve() # solves equation 2*x - 3*y - 5 == 0 {(3*t_0 - 5, 2*t_0 - 5)} Here x = -3*t_0 - 5 and y = -2*t_0 - 5 >>> Linear(2*x - 3*y - 4*z -3).solve() {(t_0, 2*t_0 + 4*t_1 + 3, -t_0 - 3*t_1 - 3)} linearc URS:H$r)rrs rDrLinear.matchess  A%%rFNcURU5 URnURnSU;aUS*nOSn[X@RS9nURn[ U5S:Xa.[ XSUS5upU (dURU45 U$UV s/sHoU PM n n /n [ U5S:aU R[U SU S55 U SU S-U S'U SU S-U S'[[ U 5S- SS5H:n [U SX5nU SU-U S'XU-X'U RSU5 M< U RU S5 /n[X5GH^unn//nn[R"U5GHnUR(aU[ R"nnUSnO)UR%5unnXwR'U5S-n[)UUUU5=nunnU[ R"La SU;aUs s $On[+U[5(a"UR,SU-UR,S-n[+U[5(a"UR,SU-UR,S-nURU5 URU5 GM UR[U65 [U6nGMa URU5 URU5 U$s sn f)Nrrr)rrrr9r@rHdivmodrOappendrrJinsertr]r make_args is_IntegerrOne as_coeff_Mulindexbase_solution_linear isinstancerQ)rAr@rrvarr~rbparamsqrrfABrRgcdrXAiBitot_xtot_yargkrepnewsolsol_xsol_ys rDr Linear.solves z"  :q AA'H"" s8q=!3q6]+DA A4 M( T # #s!1Xs #  s8a< HHT!B%2' (bEQqTMAbEbEQqTMAbE3q6A:q"-1Q4&ts{!ts{C . 23 h !iFBr5E}}Q'>>qA!!9D++-DAq!,,q/A"56D%9!RT%JJleU:s{% #"%-- % 1 a%**Q- ?!%-- % 1 a%**Q- ? U# U#1(4   S%[ )U A= @  9 I $sK8rrrrrFrDrrs2 D&drFrc0\rSrSrSrSrSrSS SjjrSrg) BinaryQuadraticia\ Representation of a binary quadratic diophantine equation. A binary quadratic diophantine equation is an equation of the form `Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0`, where `A, B, C, D, E, F` are integer constants and `x` and `y` are integer variables. Examples ======== >>> from sympy.abc import x, y >>> from sympy.solvers.diophantine.diophantine import BinaryQuadratic >>> b1 = BinaryQuadratic(x**3 + y**2 + 1) >>> b1.matches() False >>> b2 = BinaryQuadratic(x**2 + y**2 + 2*x + 2*y + 2) >>> b2.matches() True >>> b2.solve() {(-1, -1)} References ========== .. [1] Methods to solve Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0, [online], Available: https://www.alpertron.com.ar/METHODS.HTM .. [2] Solving the equation ax^2+ bxy + cy^2 + dx + ey + f= 0, [online], Available: https://web.archive.org/web/20160323033111/http://www.jpr2718.org/ax2p.pdf binary_quadraticcLURS:H=(a URS:H$Nrrrrs rDrBinaryQuadratic.matches!  A%=$..A*==rFNc d^;^<^=^>^?^@^A^B^CURU5 URnURnUupVXES-nXEU-nXFS-n XEm;XFm[UT;- U5unnU(aMYU RUU45 Mn U $U S:XGaUS:XaH[URXe/S9RTCU /S9nUHnU RUSUS45 M U $[U5[!Xy5-m@UT@-nU T@-n[X- 5m?[#U5mA[#U5mBT?TB-T;-TAT<-- m>T>(d[%SSS 9nTAT@-US--T;U--TAT=--n['UU5R)[R*5nUH6n[-TAU-T?TB-U--U- 5nU RUSUS45 M8 U $[/U5(aUU?U@UBUC4S jnU;U=U>U@UAUC4S jn[1S[3T>55Hxn [5TAT@-U S--T;U --TAT=--T>5(dM+[5T?TB-T@-U S--T5(dMYU RU"U 5U"U 545 Mz U $[7U 5(GaUS:wGa[9U 5n[;S SS 9un n![=SU-U-U -U!-SU-T;-UU!-X--UU!--X-- --SU-S-U-T<-U U!- --SU-U-S-U-T=--5n[-UTC5n"U"GHun#n$UU$-UU#--UU$--UU#-- n%[U%5SU-U-- n&[U#U$- 5SU-- n'[?U#[$5(d[?U$[$5(aH[A[CU&U'SU-U-U55S:a%[CU&U'SU-U-U5nU RD"U6 MMU&RF(dMU'RF(dM[IX4U&U'5(dMU RU&U'45 GM U $[URUSSS2S9RTCU /S9nU(a.U RURK5SSS25 U(aM.U $[MX45un(n)[OX45um;n*[QT;U*5n+T;S:adU+H\unnU*U4HMnU*U4HAnU([SXV/5-U)-nU RUV,s/sHn,[ U,5PM sn,5 MC MO M^ U $[WU+5n+U+RES[YU+555 [QT;S5nUSSn-USSn.[[SU(SSU)SS-55(aU+HunnUU[9T;5--U-U.[9T;5--TC--n/UU[9T;5-- U-U.[9T;5-- TC--n0[=[U/U0-5S- 5n1[=[U/U0- 5S[9T;5-- 5n2U([SU1U2/5-U)-nU RU5 M U $[]U(SSU)SS-V,s/sHn,U,R^PM sn,6n3Sn4U-n5U.n6U5S- U3-S:wd U6U3-S:wa7U5U--T;U6-U.--U5U.-U6U---n6n5U4S- n4U5S- U3-S:waM,U6U3-S:waM7U+Hun7n8[1U45Hn [[SU([SU7U8/5-U)-55(aU7[9T;5U8--U5[9T;5U6--TC--n/U7[9T;5U8-- U5[9T;5U6-- TC--n0[U/U0-5S- n9[U/U0- 5S[9T;5-- n:U([SU9U:/5-U)-nU RU5 U7U--T;U.-U8--U7U.-U8U---n8n7M M U $s sn fs snfs sn,f![Ta GMf=fs sn,f)NrrrrrzT)realc>T*T-T-T-TS--TST-T-T-U--T-- TT-T-US--TU--TT-T--T-- $rr)uEF_cegsqcts rD'BinaryQuadratic.solve..8sn3q AqD(8A!C ! OQ;N(N+,S571a4TT-T-TS--TST-T-U--T--TT-US--TU--TT--T--$rr)rDrrrsqars rDrr;s[AbA QsU1WQY8I(I+.q5A:!+;c!e+C*J)KrFzu, vrrc32# UH upU*U*4v M g7frWr)r}XYs rDr(BinaryQuadratic.solve..ys!H7GtqA2r(7Gsc38# UHn[U5v M g7frWr|r}rTs rDrrsrIrrr r)DrAr@rrrrSyrrCrRrbrdiscrrrdivtermdx0y0ssolnar~reqrootsrootanssolve_xsolve_yz0rfrPs0t0numx_0y_0PQN solns_pellrTTU_a_bx_ny_nLrT_kU_krrXtYtrrrrrrrrrsD @@@@@@@@@rDrBinaryQuadratic.solves  z"  Q$K A#J Q$K H H !%%L/:1Aq!/LM/L!F1I/LMaAq(__=  11qs1u  6a1fasQqSyA~a|JJAw'a|JJA2w'h eqsQqSy)s3stdUs33A"1q5!,EB1%acAaCi3 q$*1q5!$4EB#$1 & B8 4` EaZAv#DMMGMMZ[]^Y_M`DJJQa12| uGDJ&FFKAhAhsU1Ws1u_s.AQq!tac)CE1B)"a0::1::FE %(Q3q4)?@ CFCF#34!&\ U ]]OOGKKG$As2w/%c!eBEkAbD&83q5&@"EE%aeAgb!emad&:QsU1W&DbII"JJ WR['AB 0F su  AvKvt41aCE!GAI!AqsQSy1Q3'< ==aCE!GAIq1u%&()!Aa ! 45&b!,&FBB$2+",qt3CC&AaCE*CBG*!,C!"f--B1G1G{3QqSUJGH1L"-c3!Az"JC"MM3/MCNNN+CSAA"JJSz2'^ C$DMMDbD JPP]^`a\bPcJJquuwtt}-a@ s)4DAqC'DAq AJ1u(FB!c2Y#%#rA !&!. 01 4A% & q+Aq!F1Iq+A B"+')h S!_ !!!HtJ7G!HHAqMaDGaDG D>vvrFrc"\rSrSrSrSrSrSrg)InhomogeneousTernaryQuadraticizu Representation of an inhomogeneous ternary quadratic. No solver is currently implemented for this equation type. inhomogeneous_ternary_quadraticcURS:XaURS:XdgUR(dgUR(+$)NrrF)rrrrrs rDr%InhomogeneousTernaryQuadratic.matchess9!!Q&4>>Q+>))))rFrNrjrkrlrmrnrurrorrFrDr7r7s -D*rFr7c0\rSrSrSrSrSrSS SjjrSrg) !HomogeneousTernaryQuadraticNormalia1 Representation of a homogeneous ternary quadratic normal diophantine equation. Examples ======== >>> from sympy.abc import x, y, z >>> from sympy.solvers.diophantine.diophantine import HomogeneousTernaryQuadraticNormal >>> HomogeneousTernaryQuadraticNormal(4*x**2 - 5*y**2 + z**2).solve() {(1, 2, 4)} $homogeneous_ternary_quadratic_normalcd^URS:XaURS:XdgUR(dgUR(dgURVs/sHoRU(dMUPM snm[ T5S:H=(a [ U4SjUR55$s snf)NrrFc32># UH oS-T;v M g7frNrr}rRnonzeros rDr.s(TBSQABSrrrrrrHrrrArrCs @rDr)HomogeneousTernaryQuadraticNormal.matchess!!Q&4>>Q+>%%"jj:jJJqM1j:7|q TS(T$BSBS(T%TT;s B-0B-Nc URU5 URnURnUupVnXES-nXFS-n XGS-n [XU SS9uupn upnunnnU*U-nU*U-n[ X0R S9nUS:aUS:aU$[ U*U-U5b"[ U*U-U5b[ U*U-U5cU$[UU5unnn[U[U55unnUU-nUU-n[UUU5unnn[U5[U 5:Xa0[UUU[U5[U5[U55unnnOw[U5[U 5:Xa0[UUU[U5[U5[U55unnnO/[UUU[U5[U5[U55unnn[UUU5n[UUU5n[XU5n[XU 5n[UU-U -5n[UU-U -5n[UU-U -5nUR[UUU55 U$)NrT)stepsrr)rrr sqf_normalr9r@r'descent _rational_pqrrrholzer reconstructr rO)rAr@rrrrSrrrbr~sqf_of_asqf_of_bsqf_of_ca_1b_1c_1a_2b_2c_2rrrbz_0r$r%rsq_lcms rDr'HomogeneousTernaryQuadraticNormal.solvesU z" a Q$K Q$K Q$K qQd + I&X3#sCDH DH'H q5QUM cT#Xs # + cT#Xs # + cT#Xs # +M1 S#c3s8,Q q q#Cc2 S# 7d1g "3S#c(CHc#hOMCc !WQ "3S#c(CHc#hOMCc"3S#c(CHc#hOMCc#sC(#sC(#C(h(3#f*()#f*()#f*() ;sC-. rFrrrrrrFrDr=r=s  2D U99rFr=c,\rSrSrSrSrSrSSjrSrg) HomogeneousTernaryQuadraticia Representation of a homogeneous ternary quadratic diophantine equation. Examples ======== >>> from sympy.abc import x, y, z >>> from sympy.solvers.diophantine.diophantine import HomogeneousTernaryQuadratic >>> HomogeneousTernaryQuadratic(x**2 + y**2 - 3*z**2 + x*y).solve() {(-1, 2, 1)} >>> HomogeneousTernaryQuadratic(3*x**2 + y**2 - 3*z**2 + 5*x*y + y*z).solve() {(3, 12, 13)} homogeneous_ternary_quadraticcn^URS:XaURS:XdgUR(dgUR(dgURVs/sHoRU(dMUPM snm[ T5S:H=(a [ U4SjUR55(+$s snf)NrrFc32># UH oS-T;v M g7frArrBs rDr6HomogeneousTernaryQuadratic.matches..)s-YGX!dgoGXrErFrGs @rDr#HomogeneousTernaryQuadratic.matches s!!Q&4>>Q+>%%"jj:jJJqM1j:LA%Y#-YtGXGX-Y*YZZ;s B20B2NcF^URU5 URnURmUupEnXEU/n[XpRS9nSn [ U4SjU55(dTXF-(aX[ TXE-U-TXV-U--XF-- 5n [U SS9n UR[U STXF-*U S55 U$USUSsUS'US'U "[UT55upnU bURXU45 U$TUS-S:Xa\TUS-S:Xa(USUSsUS'US'U "[UT55upn GOUSUSsUS'US'U "[UT55upnGOTXE-(d TXF-(aTUS-nTXE-nTXF-nTUS-nTXV-nTUS-n0nS US--UUS-'S U-U-US-- UUS-'S U-U-US-- UUS-'S U-U-SU-U-- UXV-'SUXE-'SUXF-'U "[UU55upnU cU$[UU -UU--SU-5unnU U-U- U U-UU-pn OTXe-S:waTUS-S:XaXTUS-S:Xa%TUS-nTXV-n[U*U5unnUUUpn OcUSUSsUS'US'U "[UT55upn O[U5S:a[U5S$g)NrNNN)rHr)rs rD unpack_sol5HomogeneousTernaryQuadratic.solve..unpack_sol?s3x!|Cy|##rFc34># UH nTUS-v M g7frArr}rRrs rDr4HomogeneousTernaryQuadratic.solve..Ds,15A;c<[US5[US5-$Nrrr)rs rDr3HomogeneousTernaryQuadratic.solve..GsC!IAaD ,ArFryrrrr) rrrr9r@anyr6minrOr_diop_ternary_quadraticrM_diop_ternary_quadratic_normal)rAr@r_varrSrrrrbrgsolsrr%r$rZrrrrrr_coeffrerrPrrs @rDr!HomogeneousTernaryQuadratic.solve+s z"   aQi(H $ ,,,,QSz"5:a<%*Q,#>#DE"AB ;qteACj[!A$?@ !!Wd1gNCFCF&'>sE'JKMCc Cc?+M A;! QT{a!%a$q'AA *+B3+N O #"&a$q'AA *+B3+N O #QSzU13Z!Q$K!#J!#J!Q$K!#J!Q$K Avq!t  s1uq!t|q!t  s1uq!t|q!t c!eac!emqs qs qs *+B3+O P #;!M#AcEAcEM1Q371 #A 3q5#a%##qsqA;!#QT{a'!!Q$K!!#J+QB21()1a##*.a$q'AA(23J3PU3V(W #&*!Wd1gNCFCF$./FsE/R$SMCc!++I#u+U V # ;M ;s-. rFrrrrrFrDr^r^s  +D [hrFr^c"\rSrSrSrSrSrSrg)InhomogeneousGeneralQuadraticizu Representation of an inhomogeneous general quadratic. No solver is currently implemented for this equation type. inhomogeneous_general_quadraticcURS:XaURS:dgUR(dg[SUR55=(a UR (+$)NrrFTc38# UHoRv M g7frWis_Mulr}rs rDr8InhomogeneousGeneralQuadratic.matches..0Z88Zrrrrrqrrrs rDr%InhomogeneousGeneralQuadratic.matchessM!!Q&4>>Q+>%%0TZZ00I9I9I5IIrFrNr;rrFrDrzrzs -DJrFrzc"\rSrSrSrSrSrSrg)HomogeneousGeneralQuadraticizr Representation of a homogeneous general quadratic. No solver is currently implemented for this equation type. homogeneous_general_quadraticcURS:XaURS:dgUR(dg[SUR55=(a UR $)NrrFc38# UHoRv M g7frWr~rs rDr6HomogeneousGeneralQuadratic.matches..rrrrs rDr#HomogeneousGeneralQuadratic.matchessJ!!Q&4>>Q+>%%0TZZ00ET5E5EErFrNr;rrFrDrrs +DFrFrc,\rSrSrSrSrSrSSjrSrg) GeneralSumOfSquaresia[ Representation of the diophantine equation `x_{1}^2 + x_{2}^2 + . . . + x_{n}^2 - k = 0`. Details ======= When `n = 3` if `k = 4^a(8m + 7)` for some `a, m \in Z` then there will be no solutions. Refer [1]_ for more details. Examples ======== >>> from sympy.solvers.diophantine.diophantine import GeneralSumOfSquares >>> from sympy.abc import a, b, c, d, e >>> GeneralSumOfSquares(a**2 + b**2 + c**2 + d**2 + e**2 - 2345).solve() {(15, 22, 22, 24, 24)} By default only 1 solution is returned. Use the `limit` keyword for more: >>> sorted(GeneralSumOfSquares(a**2 + b**2 + c**2 + d**2 + e**2 - 2345).solve(limit=3)) [(15, 22, 22, 24, 24), (16, 19, 24, 24, 24), (16, 20, 22, 23, 26)] References ========== .. [1] Representing an integer as a sum of three squares, [online], Available: https://proofwiki.org/wiki/Integer_as_Sum_of_Three_Squares general_sum_of_squaresc^TRS:XaTRS:dgTR(dg[STR55(ag[ U4SjTR55$)NrrFc38# UHoRv M g7frWr~rs rDr.GeneralSumOfSquares.matches..,Axxrc3V># UHoS:wdM TRUS:Hv M g7frNrr}rrAs rDrr%Dz!!V%4::a=A%z )))rrrrqrrrs`rDrGeneralSumOfSquares.matchessX!!Q&4>>Q+>%% ,, , ,DtzzDDDrFNc ,URU5 URn[URS5*nURn[ X0R S9nUS:dUS:aU$UVs/sHowR(aSOSPM nnURS5S:gn Sn [XESS9H]n U (a5UR[U 5V V s/sH upXU -PM sn n 5 OURU 5 U S- n X:XdM\ U$ U$s snfs sn n f)NrrrrTzeros) rrrLrrr9r@is_nonpositivecountsum_of_squaresrO enumerate)rAr@rrrnrbrSsignsnegstookrrRjs rDrGeneralSumOfSquares.solves z" A   NN'H q5EAIM8;<1''Q.<{{2!#D1A 9Q<@<41EHQJ<@A 1 AID} 2 = As (D  D rrrrrFrDrrs@ $DErFrc<\rSrSrSrSrSr\S5rS Sjr Sr g) GeneralPythagoreaniar Representation of the general pythagorean equation, `a_{1}^2x_{1}^2 + a_{2}^2x_{2}^2 + . . . + a_{n}^2x_{n}^2 - a_{n + 1}^2x_{n + 1}^2 = 0`. Examples ======== >>> from sympy.solvers.diophantine.diophantine import GeneralPythagorean >>> from sympy.abc import a, b, c, d, e, x, y, z, t >>> GeneralPythagorean(a**2 + b**2 + c**2 - d**2).solve() {(t_0**2 + t_1**2 - t_2**2, 2*t_0*t_2, 2*t_1*t_2, t_0**2 + t_1**2 + t_2**2)} >>> GeneralPythagorean(9*a**2 - 4*b**2 + 16*c**2 + 25*d**2 + e**2).solve(parameters=[x, y, z, t]) {(-10*t**2 + 10*x**2 + 10*y**2 + 10*z**2, 15*t**2 + 15*x**2 + 15*y**2 + 15*z**2, 15*t*x, 12*t*y, 60*t*z)} general_pythagoreanc^TRS:XaTRS:dgTR(dg[STR55(ag[ U4SjTR55(ag[ U4SjTR55(dg[ [U4SjTR555TRS- :H$)NrrFc38# UHoRv M g7frWr~rs rDr-GeneralPythagorean.matches..rrc3V># UHoS:wdM TRUS:Hv M g7frrrs rDrrs%@:aa!tzz!}!:rc3f># UH&n[[TRU55v M( g7frW)r$rrrs rDrr s&E*Q9SA/00*s.1c3T># UHn[TRU5v M g7frW)rrrs rDrr$s!?JqtDJJqM**Js%()rrrrqrrrsumrs`rDrGeneralPythagorean.matchess!!Q&4>>Q+>%% ,, , , @4::@ @ @E$**EEE3?DJJ??@DNNUVDVVVrFc URS- $rrrs rDrGeneralPythagorean.n_parameters&s~~!!rFNc HURU5 URnURnURn[ X4SS-5[ X4SS-5-[ X4SS-5-S:aUR 5H nX6*X6'M [ X@RS9nSn[U5Hup[ X:S-5S:XdMU nM URn [SU 55n U SXS- S--- /n U R[US- 5V s/sHn SX-XS- -PM sn 5 U SUU /-XS-nSn[U5HxupX:XdUS:aU S:Xd US:Xa+U S:Xa%[U[[X:S-555nMG[X:S-5n[U[U5(aUOUS-5nMz [U5H(upXU -[[X:S-55- X'M* UR!U5 U$s sn f)Nrrrrrc3*# UH oS-v M g7frAr)r}m_is rDr+GeneralPythagorean.solve..?s(as(as)rrrrrkeysr9r@rrextendrJr rr_oddrO)rAr@rrrrrzrbrrRrfmithr0rlcmrs rDrGeneralPythagorean.solve*s z"  NN !fk" #d5Q1+=&> >ePQFVWKFXAY Y\] ]zz|#j[ $(HcNDAEq&M"b(#   (a(( 1qQx1}$ $ % uQU|<|!!ad(Q1uX%|<=i3%!F)+cNDAzeaiAF qAv3SAv%7 89Av'3T!WW!q&9 #cNDAFld3u!V}+=&>>CF#  3 =sHrr) rjrkrlrmrnrurrrrrorrFrDrrs-  !D W""&rFrc"\rSrSrSrSrSrSrg) CubicThueiSa Representation of a cubic Thue diophantine equation. A cubic Thue diophantine equation is a polynomial of the form `f(x, y) = r` of degree 3, where `x` and `y` are integers and `r` is a rational number. No solver is currently implemented for this equation type. Examples ======== >>> from sympy.abc import x, y >>> from sympy.solvers.diophantine.diophantine import CubicThue >>> c1 = CubicThue(x**3 + y**2 + 1) >>> c1.matches() True cubic_thuecLURS:H=(a URS:H$)Nrrrrs rDrCubicThue.matchesjrrFrNr;rrFrDrrSs( D>rFrc,\rSrSrSrSrSrSSjrSrg) GeneralSumOfEvenPowersinaK Representation of the diophantine equation `x_{1}^e + x_{2}^e + . . . + x_{n}^e - k = 0` where `e` is an even, integer power. Examples ======== >>> from sympy.solvers.diophantine.diophantine import GeneralSumOfEvenPowers >>> from sympy.abc import a, b >>> GeneralSumOfEvenPowers(a**4 + b**4 - (2**4 + 3**4)).solve() {(2, 3)} general_sum_of_even_powersc^TRS:dgTRS-S:wag[U4SjTR55(dg[U4SjTR55$)NrFrrc3># UH8oS:wdM UR=(a URTR:Hv M: g7fr)is_Powexprrs rDr1GeneralSumOfEvenPowers.matches..s5YJqWXRX:188:):): ::Js A3Ac3V># UHoS:wdM TRUS:Hv M g7frrrs rDrrrr)rrrrs`rDrGeneralSumOfEvenPowers.matchessW  1$   q A %YDJJYYYDtzzDDDrFNc URU5 URnURnSnUR5H-nUR(dMXF(dM!UR nM/ [ U5nUS*n[X0RS9n US:dUS:aU $UV s/sHoR(aSOSPM n n U R"S5S:gn Sn [XU5H^nU (a6U R[U5VVs/sH unnXU-PM snn5 OU RU5 U S- n X:XdM] U $ U $s sn fs snnf)Nrrrr)rrrrrrrHr9r@rrpower_representationrOr)rAr@rrrrerrrrbrSrrrrrRrs rDrGeneralSumOfEvenPowers.solves+ z"  AxxxEHHEE H 1XI'H q5EAIM7:;s!&&A-s;zz"~"%aA.A )A,?,$!QDGAI,?@ 1 AID} / < @s #E  E rrrrrFrDrrns" (DErFrc`[U6nUS:XaU$[UVs/sHo3U-PM sn5$![aZ [[SU55n[ U5S:aUs$[UVs/sHo3R 5SPM Os snfsn6nN[ a [ S5ef=fs snf)Nrrz-_remove_gcd(a,b,c) or _remove_gcd(*container)r)rr>rfilterrHas_content_primitiverr?)rSrfxrRs rDrrsI !H Av "1Q$" ## = &q/ " r7Q;H ;1))+A.; < IGHHI#s&0B+0B(" B(+B  B(B(cD[[U5U-[U55$rW)rrrrrPs rDrMrMs tAwqy#a& ))rFc`[X5up#[U5[U5S-::aU$US-$)Nrr)rr)rerwrs rD_nint_or_floorrs/ !4a [[AS 55ef=f![Ra SnGN>> from sympy import diophantine >>> from sympy.abc import a, b >>> eq = a**4 + b**4 - (2**4 + 3**4) >>> diophantine(eq) {(2, 3)} >>> diophantine(eq, permute=True) {(-3, -2), (-3, 2), (-2, -3), (-2, 3), (2, -3), (2, 3), (3, -2), (3, 2)} >>> from sympy.abc import x, y, z >>> diophantine(x**2 - y**2) {(t_0, -t_0), (t_0, t_0)} >>> diophantine(x*(2*x + 3*y - z)) {(0, n1, n2), (t_0, t_1, 2*t_0 + 3*t_1)} >>> diophantine(x**2 + 3*x*y + 4*x) {(0, n1), (-3*t_0 - 4, t_0)} See Also ======== diop_solve sympy.utilities.iterables.permute_signs sympy.utilities.iterables.signed_permutations Trwryz/syms should be given as a sequence, e.g. a list)permuteas_AddFrc38# UHoRv M g7frW) is_number)r}rs rDrdiophantine..Fs3Fq{{Frz@ Equation should be a polynomial with Rational coefficients.rrc36# UHoSUS-v M g7frrNr)r}rs rDrrl$@1qT!A$Yrc36# UHoSUS-v M g7frr)r}rSs rDrrrrr)paramsymsr_dict)evaluatec3>># UHn[TTU5v M g7frW)merge_solution)r}rrvar_ts rDrrsL8CsE3778szunhandled type: %src3V# UHup[U40URD6SLv M! g7f)FN)r assumptions0)r}valrs rDrrs(bSa 77uDSas')c38# UHn[U5v M g7frWr|)r}rs rDrrs*cz!}}crc2>USUS-TSTS-:H$rnr)rSrs rDrdiophantine..s"AaD1IQA,FrF)=rrrlhsrhsrrrrrr1rr\r]rJrHr6r?as_numer_denomrrIrrNras_independentr*rqgensas_expr is_polynomialr)AssertionErrorr0r7rrurr^r=rr2KeyErrorboolrr+ is_Rationalr,rrrrrOrrrYdiscardis_zerorr3rr4),rrrrrRdict_sym_indexrrrdsolgoodrredo_permute_signsdo_permute_signs_varpermute_few_signsrfr~len_varpermute_signs_forpermute_signs_checkvar_mulxy_coeffx_coeff var1_mul_var2 v1_mul_v2rv1termsflrvrbaserTeq_typerPnull final_soln permuted_signlstpermuted_sign_varrrrs, @@@rDr6r6sL "B"b VVbff_E29949(556 %& t$$EGG#0t!CxAtD0s{!%c$c$i0@&A!B%0G%LN%LSAS!23SAB%LNN  " ;;5L{{q>Dq>D(D#Et!xs3{0C'DAtE E !_<<    2E 21 5 H3AFF33333 YY[!!!!  X #1a !fG#((&++!- ,00166$$#& %%'+$))a<"71a=1G#H"G$@$@M&3 &$%iLE$(>#Ad5k &3 #71a=1G%&$%beHE#'w-"?DK & '233+/($,0)\"71a=1G#H"G$@$@M%2 &$%iLE$(>#Ad5k &3 #71a=1G%&$%beHE#'w-"?DK & '233,0($ -1) ( (!WIEO 5Da)$e<q'4%0==?4dE*  +00166"'' ) ) HH^C9 : $$#((&++ !! KKL8L L&&:W&DE E-0 LL !SX D BGGCTN+33 bSVW[]`Sab b b HHTNJ *c* * * #M#$6 7 !!-0"=-.6"FLM #C !!-0%$'(;C(@$A!!!"34s#s#!" U1BNF n -E $CDE EET (&$%E& (&$%E&" (&$%E& (&$%E&& * + _ a5  A!r"Q%xuQ Q1 sA[5 [ &[ ,<[5([*7[% [*[5,[5.[51-[0"[0([5+B[5;B/])+\0;]),\04A-])"]';])#]+A ])6]) [5%[** [55%\ \-)]),\--])0 ]<])?]]) ]])]]) ]&"])%]&&]))A ^=4^=<^=c</nSU;ag[U5n[SSSS9nUH>nXQ;aUR[U55 M$UR[U55 M@ [ X05H upg[ U40UR D6SLdM g [U5$)a This is used to construct the full solution from the solutions of sub equations. Explanation =========== For example when solving the equation `(x - y)(x^2 + y^2 - z^2) = 0`, solutions for each of the equations `x - y = 0` and `x^2 + y^2 - z^2` are found independently. Solutions for `x - y = 0` are `(x, y) = (t, t)`. But we should introduce a value for z when we output the solution for the original equation. This function converts `(t, t)` into `(t, t, n_{1})` where `n_{1}` is an integer parameter. NrrTr)rstartF)iterr.rnextr]rrr?)rrrPrrrfrsymbs rDrrs C xH~H c4q 9F  : JJtH~ & JJtF| $  ]  S 6D$5$5 6% ?# :rFc[H5nU"U5R5(dM U"U5RUS9s $ g)Nr)all_diop_classesrr)rr diop_types rD _diop_solver&s7% R= " "R=&&&&9 9&rFc[USS9up#nU[R:Xa [X5$U[R:Xa [ X5$U[ R:Xa [USS9$U[R:Xa [USS9$U[R:Xa [X5$U[R:Xa [U5$U[R:Xa[U[ R"S9$U[$R:Xa['U[ R"S9$UbU[(;a[+[-S55e[/SU-5e)a Solves the diophantine equation ``eq``. Explanation =========== Unlike ``diophantine()``, factoring of ``eq`` is not attempted. Uses ``classify_diop()`` to determine the type of the equation and calls the appropriate solver function. Use of ``diophantine()`` is recommended over other helper functions. ``diop_solve()`` can return either a set or a tuple depending on the nature of the equation. All non-trivial solutions are returned: assumptions on symbols are ignored. Usage ===== ``diop_solve(eq, t)``: Solve diophantine equation, ``eq`` using ``t`` as a parameter if needed. Details ======= ``eq`` should be an expression which is assumed to be zero. ``t`` is a parameter to be used in the solution. Examples ======== >>> from sympy.solvers.diophantine import diop_solve >>> from sympy.abc import x, y, z, w >>> diop_solve(2*x + 3*y - 5) (3*t_0 - 5, 5 - 2*t_0) >>> diop_solve(4*x + 3*y - 4*z + 5) (t_0, 8*t_0 + 4*t_1 + 5, 7*t_0 + 3*t_1 + 5) >>> diop_solve(x + 3*y - 4*z + w - 6) (t_0, t_0 + t_1, 6*t_0 + 5*t_1 + 4*t_2 - 6, 5*t_0 + 4*t_1 + 3*t_2 - 6) >>> diop_solve(x**2 + y**2 - 5) {(-2, -1), (-2, 1), (-1, -2), (-1, 2), (1, -2), (1, 2), (2, -1), (2, 1)} See Also ======== diophantine() FrT) parameterizerz Although this type of equation was identified, it is not yet handled. It should, however, be listed in `diop_known` at the top of this file. Developers should see comments at the end of `classify_diop`. r)r7rru diop_linearrdiop_quadraticr^diop_ternary_quadraticr=diop_ternary_quadratic_normalrdiop_general_pythagoreanrdiop_univariaterdiop_general_sum_of_squaresrInfinityrdiop_general_sum_of_even_powers diop_knownr>r0r)rrrrrs rDrr s,`(%8C&++2%% O(( (b(( /44 4%bt<< 5:: :,RdCC &++ +'22 JOO #r"" ',, ,*2QZZ@@ *// /.rDDwj8Z) " 07 :< >> from sympy.solvers.diophantine import classify_diop >>> from sympy.abc import x, y, z, w, t >>> classify_diop(4*x + 6*y - 4) ([x, y], {1: -4, x: 4, y: 6}, 'linear') >>> classify_diop(x + 3*y -4*z + 5) ([x, y, z], {1: 5, x: 1, y: 3, z: -4}, 'linear') >>> classify_diop(x**2 + y**2 - x*y + x + 5) ([x, y], {1: 5, x: 1, x**2: 1, y**2: 1, x*y: -1}, 'binary_quadratic') z * cl[USS9up#nU[R:XaSnUb[SU[ U54-SS9n[U5R US9nUcU"S/[ UR 5-6n[ U5S:a[U5S$[S/[ UR 5-5$g) a Solves linear diophantine equations. A linear diophantine equation is an equation of the form `a_{1}x_{1} + a_{2}x_{2} + .. + a_{n}x_{n} = 0` where `a_{1}, a_{2}, ..a_{n}` are integer constants and `x_{1}, x_{2}, ..x_{n}` are integer variables. Usage ===== ``diop_linear(eq)``: Returns a tuple containing solutions to the diophantine equation ``eq``. Values in the tuple is arranged in the same order as the sorted variables. Details ======= ``eq`` is a linear diophantine equation which is assumed to be zero. ``param`` is the parameter to be used in the solution. Examples ======== >>> from sympy.solvers.diophantine.diophantine import diop_linear >>> from sympy.abc import x, y, z >>> diop_linear(2*x - 3*y - 5) # solves equation 2*x - 3*y - 5 == 0 (3*t_0 - 5, 2*t_0 - 5) Here x = -3*t_0 - 5 and y = -2*t_0 - 5 >>> diop_linear(2*x - 3*y - 4*z -3) (t_0, 2*t_0 + 4*t_1 + 3, -t_0 - 3*t_1 - 3) See Also ======== diop_quadratic(), diop_ternary_quadratic(), diop_general_pythagorean(), diop_general_sum_of_squares() FrNz%s_0:%iTrrr) r7rrurrHrr@rr?)rrrrr%r@rbs rDr*r*sP*"E:C FKK   eSX->!>MJ!!Z!8 =aSV%6%6!778F v;?<? "$F$5$5 667 7 rFc4[XU5upnUS:XaUcgUS:aU*nX#-U*U-4$[[U5[U55upEnU[U5-nU[U5-nX-(agUcX-X-4$US:aU*nX-X#--X-X-- 4$)a, Return the base solution for the linear equation, `ax + by = c`. Explanation =========== Used by ``diop_linear()`` to find the base solution of a linear Diophantine equation. If ``t`` is given then the parametrized solution is returned. Usage ===== ``base_solution_linear(c, a, b, t)``: ``a``, ``b``, ``c`` are coefficients in `ax + by = c` and ``t`` is the parameter to be used in the solution. Examples ======== >>> from sympy.solvers.diophantine.diophantine import base_solution_linear >>> from sympy.abc import t >>> base_solution_linear(5, 2, 3) # equation 2*x + 3*y = 5 (-5, 5) >>> base_solution_linear(0, 5, 7) # equation 5*x + 7*y = 0 (0, 0) >>> base_solution_linear(5, 2, 3, t) # equation 2*x + 3*y = 5 (3*t - 5, 5 - 2*t) >>> base_solution_linear(0, 5, 7, t) # equation 5*x + 7*y = 0 (7*t, -5*t) rrrr)rr rr)r~rrPrrrrs rDrrs>!"GA!Av 9 q5AaRT{s1vs1v&IBA$q'MB$q'MBuyad|1u B D13Jqs ##rFc[USS9upnU[R:XaF[XS5R [ R 5Vs1sHn[U54iM sn$gs snf)a Solves a univariate diophantine equations. Explanation =========== A univariate diophantine equation is an equation of the form `a_{0} + a_{1}x + a_{2}x^2 + .. + a_{n}x^n = 0` where `a_{1}, a_{2}, ..a_{n}` are integer constants and `x` is an integer variable. Usage ===== ``diop_univariate(eq)``: Returns a set containing solutions to the diophantine equation ``eq``. Details ======= ``eq`` is a univariate diophantine equation which is assumed to be zero. Examples ======== >>> from sympy.solvers.diophantine.diophantine import diop_univariate >>> from sympy.abc import x >>> diop_univariate((x - 2)*(x - 3)**2) # solves equation (x - 2)*(x - 3)**2 == 0 {(2,), (3,)} FrrN)r7rrur-rrrrL)rrrr%rRs rDr/r/sq>*"E:C JOO##0 A$! !**-$./$.aQ $./ /$/sA)cX-(+$)zF Returns `True` if ``a`` is divisible by ``b`` and `False` otherwise. rrs rDrr<su9rFc[USS9up#nU[R:Xa3Ub U[SSS9/nOSn[ [U5R US95$g)a Solves quadratic diophantine equations. i.e. equations of the form `Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0`. Returns a set containing the tuples `(x, y)` which contains the solutions. If there are no solutions then `(None, None)` is returned. Usage ===== ``diop_quadratic(eq, param)``: ``eq`` is a quadratic binary diophantine equation. ``param`` is used to indicate the parameter to be used in the solution. Details ======= ``eq`` should be an expression which is assumed to be zero. ``param`` is a parameter to be used in the solution. Examples ======== >>> from sympy.abc import x, y, t >>> from sympy.solvers.diophantine.diophantine import diop_quadratic >>> diop_quadratic(x**2 + y**2 + 2*x + 2*y + 2, t) {(-1, -1)} References ========== .. [1] Methods to solve Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0, [online], Available: https://www.alpertron.com.ar/METHODS.HTM .. [2] Solving the equation ax^2+ bxy + cy^2 + dx + ey + f= 0, [online], Available: https://web.archive.org/web/20160323033111/http://www.jpr2718.org/ax2p.pdf See Also ======== diop_linear(), diop_ternary_quadratic(), diop_general_sum_of_squares(), diop_general_pythagorean() FrNrTrr)r7rrurrIr)rrrrr%r@s rDr+r+CscV*"E:C O(((  T!:;JJ?2&,, ,CDD )rFc [[XU455n[UR5VVs/sHupVXeR U5-PM snn6n[ U5S:H$s snnf)z Check whether `(u, v)` is solution to the quadratic binary diophantine equation with the variable list ``var`` and coefficient dictionary ``coeff``. Not intended for use by normal users. r)r\r]ritemsxreplacer)rrrrfrepsrRrrs rDr r xsX CQ !D ekkm 0, D` is not a perfect square, which is the same as the generalized Pell equation. The LMM algorithm [1]_ is used to solve this equation. Returns one solution tuple, (`x, y)` for each class of the solutions. Other solutions of the class can be constructed according to the values of ``D`` and ``N``. Usage ===== ``diop_DN(D, N, t)``: D and N are integers as in `x^2 - Dy^2 = N` and ``t`` is the parameter to be used in the solutions. Details ======= ``D`` and ``N`` correspond to D and N in the equation. ``t`` is the parameter to be used in the solutions. Examples ======== >>> from sympy.solvers.diophantine.diophantine import diop_DN >>> diop_DN(13, -4) # Solves equation x**2 - 13*y**2 = -4 [(3, 1), (393, 109), (36, 10)] The output can be interpreted as follows: There are three fundamental solutions to the equation `x^2 - 13y^2 = -4` given by (3, 1), (393, 109) and (36, 10). Each tuple is in the form (x, y), i.e. solution (3, 1) means that `x = 3` and `y = 1`. >>> diop_DN(986, 1) # Solves equation x**2 - 986*y**2 = 1 [(49299, 1570)] See Also ======== find_DN(), diop_bf_DN() References ========== .. [1] Solving the generalized Pell equation x**2 - D*y**2 = N, John P. Robertson, July 31, 2004, Pages 16 - 17. [online], Available: https://web.archive.org/web/20160323033128/http://www.jpr2718.org/pell.pdf rr9T generatorrrrF) all_roots)r square_factor cornacchiarLrrrJrrr>_special_diop_DNrPQar!rr'r&lengthr )rr(rrrrSrsN_exactsDsqpqarTprev_Bprev_Grr_B_Gframsqmrrs rDr r sl 1u 68O q5I-*d;A"1c1"gs119~> AC:&7JJQSz*?<  Av q5I 6F8O$Q*  G9  !A&JB 6T1I; uT!Wa!e_ad34q89A ,Q!tVaZ; v B7# : 1a4|!|%%Avx C 1v{!Ql!#YF%.s^ !A!Aq"bAbDyF&4 q5Av F+,J 7$% %qAcJQBRz mA&$ 7 AI VArT2C!#r*AaQ-C!%c Q6!R+a/0#'9 1q"bq6Q;qy1QY;.!3 AfHah#78%an,, Avad1g~1a8H'H$I$%vad1g~qtAw'F$G$IJ!#R1 38" Ja  s"M M-,M-c[U5n[[[U55SS9Vs0sH o1US--U_M nnSnSnSupxSup/n [ S5HZn XR-U-n X-U- nXS-- U-nXU-U-pXU -U -pUS-X S--- =o;dM?XNnU R X8-X:-45 M\ US:XaU $Mts snf)a Solves the equation `x^2 - Dy^2 = N` for the special case where `1 < N**2 < D` and `D` is not a perfect square. It is better to call `diop_DN` rather than this function, as the former checks the condition `1 < N**2 < D`, and calls the latter only if appropriate. Usage ===== WARNING: Internal method. Do not call directly! ``_special_diop_DN(D, N)``: D and N are integers as in `x^2 - Dy^2 = N`. Details ======= ``D`` and ``N`` correspond to D and N in the equation. Examples ======== >>> from sympy.solvers.diophantine.diophantine import _special_diop_DN >>> _special_diop_DN(13, -3) # Solves equation x**2 - 13*y**2 = -3 [(7, 2), (137, 38)] The output can be interpreted as follows: There are two fundamental solutions to the equation `x^2 - 13y^2 = -3` given by (7, 2) and (137, 38). Each tuple is in the form (x, y), i.e. solution (7, 2) means that `x = 7` and `y = 2`. >>> _special_diop_DN(2445, -20) # Solves equation x**2 - 2445*y**2 = -20 [(445, 9), (17625560, 356454), (698095554475, 14118073569)] See Also ======== diop_DN() References ========== .. [1] Section 4.4.4 of the following book: Quadratic Diophantine Equations, T. Andreescu and D. Andrica, Springer, 2015. TrBrrr)rr)rr)rr rErrJr)rr(sqrt_DrSrr&r'G0G1B0B1rXrTrrs rDrGrGsl1XF' c!f(=NON!adANAO A A FB FBI qA!AaATaA22UQ1uW_$*D  !$. 6    PsCrc[5nX-U:a}US:Xa:[X -S5upEU(aUR[U5S45 X:XaU$X!-S:Xa3[X!-S5upEU(aURS[U545 U$[ U*[ X5-U5HnXbS-:aM X&pX US--- =n S::aXU-pX US--- =n S::aM[ X5upU (aML[U S5upEU(dMcX:XaX:aXHpHUR[U5[U545 M U$)a Solves `ax^2 + by^2 = m` where `\gcd(a, b) = 1 = gcd(a, m)` and `a, b > 0`. Explanation =========== Uses the algorithm due to Cornacchia. The method only finds primitive solutions, i.e. ones with `\gcd(x, y) = 1`. So this method cannot be used to find the solutions of `x^2 + y^2 = 20` since the only solution to former is `(x, y) = (4, 2)` and it is not primitive. When `a = b`, only the solutions with `x \leq y` are found. For more details, see the References. Examples ======== >>> from sympy.solvers.diophantine.diophantine import cornacchia >>> cornacchia(2, 3, 35) # equation 2x**2 + 3y**2 = 35 {(2, 3), (4, 1)} >>> cornacchia(1, 1, 25) # equation x**2 + y**2 = 25 {(4, 3)} References =========== .. [1] A. Nitaj, "L'algorithme de Cornacchia" .. [2] Solving the diophantine equation ax**2 + by**2 = m by Cornacchia's method, [online], Available: http://www.numbertheory.org/php/cornacchia.html See Also ======== sympy.utilities.iterables.signed_permutations rrr)rIrrOrLr(rr) rrPrrvrrKrrrm1_rs rDrFrF]s:H 5Duqy 6afa(IA#a&!%v 5A:afa(IA!SV% A2fQl?A . Av: 11a4ZrA%!eq1a4ZrA% "aL  6v!%1 HHc!fc!f% &/ KrFc## [U5nS=pES=pgUnU*n Un Un X-U -n X-U-UpFX-U-UpuX-U -UpXXXX4v X-U - n X*S-- U -n M97f)ai Returns useful information needed to solve the Pell equation. Explanation =========== There are six sequences of integers defined related to the continued fraction representation of `\\frac{P + \sqrt{D}}{Q}`, namely {`P_{i}`}, {`Q_{i}`}, {`a_{i}`},{`A_{i}`}, {`B_{i}`}, {`G_{i}`}. ``PQa()`` Returns these values as a 6-tuple in the same order as mentioned above. Refer [1]_ for more detailed information. Usage ===== ``PQa(P_0, Q_0, D)``: ``P_0``, ``Q_0`` and ``D`` are integers corresponding to `P_{0}`, `Q_{0}` and `D` in the continued fraction `\\frac{P_{0} + \sqrt{D}}{Q_{0}}`. Also it's assumed that `P_{0}^2 == D mod(|Q_{0}|)` and `D` is square free. Examples ======== >>> from sympy.solvers.diophantine.diophantine import PQa >>> pqa = PQa(13, 4, 5) # (13 + sqrt(5))/4 >>> next(pqa) # (P_0, Q_0, a_0, A_0, B_0, G_0) (13, 4, 3, 3, 1, -1) >>> next(pqa) # (P_1, Q_1, a_1, A_1, B_1, G_1) (-1, 1, 1, 4, 1, 3) References ========== .. [1] Solving the generalized Pell equation x^2 - Dy^2 = N, John P. Robertson, July 31, 2004, Pages 4 - 8. https://web.archive.org/web/20160323033128/http://www.jpr2718.org/pell.pdf rrr)r) P_0Q_0rsqDA2r[A1B2rYG2P_iQ_ia_is rDrHrHsJ (CKBKB B B C C yS "bB"bB"bB''gm6zc! sAAc  [U5n[U5n/n[US5nUSSnUS:Xa:US:aS/$US:XaSU4/$[US5upgU(aXb-U4U*U-U4/$S/$[U5S:Xa [X5$US:a)Sn[[ XS- -SU-- 5S5SS-n OM[[ X- 5*S5upU(dUS- n[[ XS--SU-- 5*S5SS-n [ X5H`n [XU S---S5upU(dM#URW U 45 [XU *XU5(aMLURU *U 45 Mb U$![ a SnNZf=f)aC Uses brute force to solve the equation, `x^2 - Dy^2 = N`. Explanation =========== Mainly concerned with the generalized Pell equation which is the case when `D > 0, D` is not a perfect square. For more information on the case refer [1]_. Let `(t, u)` be the minimal positive solution of the equation `x^2 - Dy^2 = 1`. Then this method requires `\sqrt{\\frac{\mid N \mid (t \pm 1)}{2D}}` to be small. Usage ===== ``diop_bf_DN(D, N, t)``: ``D`` and ``N`` are coefficients in `x^2 - Dy^2 = N` and ``t`` is the parameter to be used in the solutions. Details ======= ``D`` and ``N`` correspond to D and N in the equation. ``t`` is the parameter to be used in the solutions. Examples ======== >>> from sympy.solvers.diophantine.diophantine import diop_bf_DN >>> diop_bf_DN(13, -4) [(3, 1), (-3, 1), (36, 10)] >>> diop_bf_DN(986, 1) [(49299, 1570)] See Also ======== diop_DN() References ========== .. [1] Solving the generalized Pell equation x**2 - D*y**2 = N, John P. Robertson, July 31, 2004, Page 15. https://web.archive.org/web/20160323033128/http://www.jpr2718.org/pell.pdf rrr9rF) r/r rrrLrJr>r equivalent) rr(rrrrrLrKL1L2rrSs rD diop_bf_DNrnsZ q Aq A C1 A !QAAv q58O 6F8O$Q*  T1IAqz* *x 1v{q}1u  SEAaC11 5a 81 <$c!#hY2  !GB c!U)QqS/22A 6q 9A = 2] 'adF A6IA 6 JJ1v aQBa00 QB7# J F sE44 FFcb[X-XA-U-- U5=(a [X-X-- U5$)a Returns True if two solutions `(u, v)` and `(r, s)` of `x^2 - Dy^2 = N` belongs to the same equivalence class and False otherwise. Explanation =========== Two solutions `(u, v)` and `(r, s)` to the above equation fall to the same equivalence class iff both `(ur - Dvs)` and `(us - vr)` are divisible by `N`. See reference [1]_. No test is performed to test whether `(u, v)` and `(r, s)` are actually solutions to the equation. User should take care of this. Usage ===== ``equivalent(u, v, r, s, D, N)``: `(u, v)` and `(r, s)` are two solutions of the equation `x^2 - Dy^2 = N` and all parameters involved are integers. Examples ======== >>> from sympy.solvers.diophantine.diophantine import equivalent >>> equivalent(18, 5, -18, -5, 13, -1) True >>> equivalent(3, 1, -18, 393, 109, -4) False References ========== .. [1] Solving the generalized Pell equation x**2 - D*y**2 = N, John P. Robertson, July 31, 2004, Page 12. https://web.archive.org/web/20160323033128/http://www.jpr2718.org/pell.pdf )r)rrfrrrr(s rDrkrk5 s2H QS13q5[! $ @139a)@@rFcSSKJn U"XU5n[US[5(a [ US5n[ U5S- nXV-$Sn[ U5nXV-$)a Returns the (length of aperiodic part + length of periodic part) of continued fraction representation of `\\frac{P + \sqrt{D}}{Q}`. It is important to remember that this does NOT return the length of the periodic part but the sum of the lengths of the two parts as mentioned above. Usage ===== ``length(P, Q, D)``: ``P``, ``Q`` and ``D`` are integers corresponding to the continued fraction `\\frac{P + \sqrt{D}}{Q}`. Details ======= ``P``, ``D`` and ``Q`` corresponds to P, D and Q in the continued fraction, `\\frac{P + \sqrt{D}}{Q}`. Examples ======== >>> from sympy.solvers.diophantine.diophantine import length >>> length(-2, 4, 5) # (-2 + sqrt(5))/4 3 >>> length(-5, 4, 17) # (-5 + sqrt(17))/4 4 See Also ======== sympy.ntheory.continued_fraction.continued_fraction_periodic r)continued_fraction_periodicrr) sympy.ntheory.continued_fractionrqrrrH)r&r'rrqrfrptnonrpts rDrIrI\ sbDM#A!,A!B%!B%jQ! <Q <rFc\[USS9upnU[R:Xa [X5$g)ac This function transforms general quadratic, `ax^2 + bxy + cy^2 + dx + ey + f = 0` to more easy to deal with `X^2 - DY^2 = N` form. Explanation =========== This is used to solve the general quadratic equation by transforming it to the latter form. Refer to [1]_ for more detailed information on the transformation. This function returns a tuple (A, B) where A is a 2 X 2 matrix and B is a 2 X 1 matrix such that, Transpose([x y]) = A * Transpose([X Y]) + B Usage ===== ``transformation_to_DN(eq)``: where ``eq`` is the quadratic to be transformed. Examples ======== >>> from sympy.abc import x, y >>> from sympy.solvers.diophantine.diophantine import transformation_to_DN >>> A, B = transformation_to_DN(x**2 - 3*x*y - y**2 - 2*y + 1) >>> A Matrix([ [1/26, 3/26], [ 0, 1/13]]) >>> B Matrix([ [-6/13], [-4/13]]) A, B returned are such that Transpose((x y)) = A * Transpose((X Y)) + B. Substituting these values for `x` and `y` and a bit of simplifying work will give an equation of the form `x^2 - Dy^2 = N`. >>> from sympy.abc import X, Y >>> from sympy import Matrix, simplify >>> u = (A*Matrix([X, Y]) + B)[0] # Transformation for x >>> u X/26 + 3*Y/26 - 6/13 >>> v = (A*Matrix([X, Y]) + B)[1] # Transformation for y >>> v Y/13 - 4/13 Next we will substitute these formulas for `x` and `y` and do ``simplify()``. >>> eq = simplify((x**2 - 3*x*y - y**2 - 2*y + 1).subs(zip((x, y), (u, v)))) >>> eq X**2/676 - Y**2/52 + 17/13 By multiplying the denominator appropriately, we can get a Pell equation in the standard form. >>> eq * 676 X**2 - 13*Y**2 + 884 If only the final equation is needed, ``find_DN()`` can be used. See Also ======== find_DN() References ========== .. [1] Solving the equation ax^2 + bxy + cy^2 + dx + ey + f = 0, John P.Robertson, May 8, 2003, Page 7 - 11. https://web.archive.org/web/20160323033111/http://www.jpr2718.org/ax2p.pdf FrN)r7rrur rrrr%s rDtransformation_to_DNrw s4\*"E:C O((($S00)rFcUup#XS-nXU-nXS-nXnXnUSn [XEXgX5V s/sHn [U 5PM sn upEpgp[SSS9upU(a[SU-U5up[XMS-5unnU S-X-X-SU S-XU-XS--- -XU-XU-U-UU-U-- SU U-U -0n[ X/U5unn[ SS[ RU - [ U5*U - SS/5U-[ SS[ RU - [ U5*U - SS/5U-4$U(a[SU-U5up[XMS-5unnU S-XU -SU S-UU-U SXU-SU U-XS--- 0n[ X/U5unn[ SS[ RU - SSS/5U-[ SS[ RU - SSS/5U-[ [ U5*U - S/5-4$U(a[SU-U5up[XmS-5unnU S-UU-X-SU S-XSU SSU U-XS--- 0n[ X/U5unn[ SSSSS[ RU - /5U-[ SSSSS[ RU - /5U-[ S[ U5*U - /5-4$[ SS[ RU- SSS/5[ SS/54$s sn f)NrrX, YTrr)rr/rrMr rrr)rrrSrrrPr~rrrSrRrrrrrr*A_0B_0s rDr r  sq DA d A c A d A A A aA+6qQ1+HI+Haq +HIA! 64 (DAAaC#A!t$1AqsACAqD!qS1T6\*:AsAs1uqQRsSTu}VWYZ[\Y\]^Y^_(!7SaQUU1WqteAgq!45c96!QqSTUVSWRWXYRY[\^_H`;abe;eeeAaC#A!t$1AqA#q!Q$!Q1c1acAdFlK(!7SaQUU1WaA./3VAq1557AqRSBT5UVY5Y\befgheidijkdkmnco\p5pppAaC#A!t$1AqsACAqD!1aAaC!qD&LI(!7SaQ1aeeAg./3VAq1aAEERSGBT5UVY5Y\bdehijkhlglmngnco\p5ppp !Qq!Q* +VQF^ ;;CJsK*c\[USS9upnU[R:Xa [X5$g)a This function returns a tuple, `(D, N)` of the simplified form, `x^2 - Dy^2 = N`, corresponding to the general quadratic, `ax^2 + bxy + cy^2 + dx + ey + f = 0`. Solving the general quadratic is then equivalent to solving the equation `X^2 - DY^2 = N` and transforming the solutions by using the transformation matrices returned by ``transformation_to_DN()``. Usage ===== ``find_DN(eq)``: where ``eq`` is the quadratic to be transformed. Examples ======== >>> from sympy.abc import x, y >>> from sympy.solvers.diophantine.diophantine import find_DN >>> find_DN(x**2 - 3*x*y - y**2 - 2*y + 1) (13, -884) Interpretation of the output is that we get `X^2 -13Y^2 = -884` after transforming `x^2 - 3xy - y^2 - 2y + 1` using the transformation returned by ``transformation_to_DN()``. See Also ======== transformation_to_DN() References ========== .. [1] Solving the equation ax^2 + bxy + cy^2 + dx + ey + f = 0, John P.Robertson, May 8, 2003, Page 7 - 11. https://web.archive.org/web/20160323033111/http://www.jpr2718.org/ax2p.pdf FrN)r7rrur rvs rDfind_DNr} s4N*"E:C O(((##)rFc Uup#[SSS9upE[X5upgU[XE/5-U-SnU[XE/5-U-Sn US-XS--X#-XU---US-XS---X!U--X1U--US-n [U R [ X#4X4555n U R 5nXS-*XS-- US*XS-- 4$)NryTrrrr)rr rrrNr]r) rrrSrrrrrrrfr simplifieds rDr r 7 s DA 64 (DA  ,DA 61&> A q!A 61&> A q!A AeqDk ACc N *QT%1+-= =( JQUVxZ WZ_`aZb bB"''#qfqf"567J  + + -E Q$K<d #eAhYuT{%: ::rFcSSKJn UR(aUR(d [ X/US9$UR(aUR(d [ X/US9$[ SSS9upVXP-Xa--R 5upxX'R-(a [ X/US9$U"X5SU"X5S- n U R 5up[XS9$) z If there is a number modulo ``a`` such that ``x`` and ``y`` are both integers, then return a parametric representation for ``x`` and ``y`` else return (None, None). Here ``x`` and ``y`` are functions of ``t``. r)clear_coefficientsrzm, nTrr)r) sympy.simplify.simplifyrrrr9rrrr&) rSrrrrrrr~rerjunks rDrrH s;{{1<<%qf@@{{1<<%qf@@ 64 (DA C!#I + + -DA33w%qf@@ A !! $'9!'?'B BB&&(HD r ))rFc[USS9up#nU[R[R4;aJ[ X#5n[ U5S:a[ U5SupgnOSupgnU(a[XgU4X#5$XgU4$g)aT Solves the general quadratic ternary form, `ax^2 + by^2 + cz^2 + fxy + gyz + hxz = 0`. Returns a tuple `(x, y, z)` which is a base solution for the above equation. If there are no solutions, `(None, None, None)` is returned. Usage ===== ``diop_ternary_quadratic(eq)``: Return a tuple containing a basic solution to ``eq``. Details ======= ``eq`` should be an homogeneous expression of degree two in three variables and it is assumed to be zero. Examples ======== >>> from sympy.abc import x, y, z >>> from sympy.solvers.diophantine.diophantine import diop_ternary_quadratic >>> diop_ternary_quadratic(x**2 + 3*y**2 - z**2) (1, 0, 1) >>> diop_ternary_quadratic(4*x**2 + 5*y**2 - z**2) (1, 0, 2) >>> diop_ternary_quadratic(45*x**2 - 7*y**2 - 8*x*y - z**2) (28, 45, 105) >>> diop_ternary_quadratic(x**2 - 49*y**2 - z**2 + 13*z*y -8*x*y) (9, 1, 5) FrrrfN)r7r^rur=rsrHr_parametrize_ternary_quadratic rr(rrr%rr$r%rZs rDr,r,d sD*"E:C  ' , , - 2 244&c1 s8a< IaLMCc,MCc 13- -}4rFc^[U4SjT55n[U5R5(a[X S9R5$[ U5R5(a[ X S9R5$g)Nc32># UH oTU-v M g7frWrrjs rDr*_diop_ternary_quadratic.. s 'AuQxZrEr)rr^rrr=)rurrs ` rDrsrs sh ' ' 'B"2&..00*2AGGII *2 . 6 6 8 80GMMOO 9rFc@[USS9upnUS;a [X5$g)a$ Returns the transformation Matrix that converts a general ternary quadratic equation ``eq`` (`ax^2 + by^2 + cz^2 + dxy + eyz + fxz`) to a form without cross terms: `ax^2 + by^2 + cz^2 = 0`. This is not used in solving ternary quadratics; it is only implemented for the sake of completeness. Frr_r>N)r7_transformation_to_normalrvs rDtransformation_to_normalr s4*"E:C 44)444rFc ^[U5nUup4n[U4SjU55(dkTX4-nTXE-nTX5-nSn U(dSn Xvpv[SSU*U- 4SSU*U- 4S45n U (a$U RSS5 U R SS5 U $TUS-S:XaTUS-S:XaCUSUSsUS'US'[ UT5n U RSS5 U R SS5 U $USUSsUS'US'[ UT5n U RSS5 U R SS5 U $TX4-S:wd TX5-S:waTUS-n TX4-n TX5-n TUS-nTXE-nTUS-n0nS U S--UUS-'S U -U-U S-- UUS-'S U -U-U S-- UUS-'S U -U-SU -U -- UXE-'SUX4-'SUX5-'[ UU5n[S S S[ U *5SU -- [ U *5SU -- SSSSSS/ 5U-$TXE-S:waTUS-S:Xa^TUS-S:Xa[S S /S Q5$USUSsUS'US'[ UT5n U RSS5 U R SS5 U $USUSsUS'US'[ UT5n U RSS5 U R SS5 U $[R"S 5$) Nc34># UH nTUS-v M g7frArrjs rDr,_transformation_to_normal.. s(CquQT{CrlFTrrrrrrrrr) rrrrrrrrr)rrqrrow_swapcol_swaprreye)rrrurSrrrrPr~swapr*rrrrrrrwT_0s ` rDrr sv 9DGA! (C( ( ( !#J !#J !#JDq QA2a4L1b1"Q$-; <  JJq!  JJq!  QT{a A;! "1vs1v DGT!W)$6A JJq!  JJq! Hq63q6Qa %dE 2 1a 1a QSzQ%*/ !Q$K !#J !#J !Q$K !#J !Q$KAvq!t s1uq!t|q!t s1uq!t|q!t c!eac!emqs qs qs 'f5aQ1"qs QrUAaC[!Q1aKLSPP qsq A;! QT{aa$@AA #1vs1v DGT!W)$6A JJq!  JJq! Hq63q6Qa %dE 2 1a 1a ::a=rFcz[USS9upnUS;a([[X55SupEn[XEU4X5$g)a4 Returns the parametrized general solution for the ternary quadratic equation ``eq`` which has the form `ax^2 + by^2 + cz^2 + fxy + gyz + hxz = 0`. Examples ======== >>> from sympy import Tuple, ordered >>> from sympy.abc import x, y, z >>> from sympy.solvers.diophantine.diophantine import parametrize_ternary_quadratic The parametrized solution may be returned with three parameters: >>> parametrize_ternary_quadratic(2*x**2 + y**2 - 2*z**2) (p**2 - 2*q**2, -2*p**2 + 4*p*q - 4*p*r - 4*q**2, p**2 - 4*p*q + 2*q**2 - 4*q*r) There might also be only two parameters: >>> parametrize_ternary_quadratic(4*x**2 + 2*y**2 - 3*z**2) (2*p**2 - 3*q**2, -4*p**2 + 12*p*q - 6*q**2, 4*p**2 - 8*p*q + 6*q**2) Notes ===== Consider ``p`` and ``q`` in the previous 2-parameter solution and observe that more than one solution can be represented by a given pair of parameters. If `p` and ``q`` are not coprime, this is trivially true since the common factor will also be a common factor of the solution values. But it may also be true even when ``p`` and ``q`` are coprime: >>> sol = Tuple(*_) >>> p, q = ordered(sol.free_symbols) >>> sol.subs([(p, 3), (q, 2)]) (6, 12, 12) >>> sol.subs([(q, 1), (p, 1)]) (-1, 2, 2) >>> sol.subs([(q, 0), (p, 1)]) (2, -4, 4) >>> sol.subs([(q, 1), (p, 0)]) (-3, -6, 6) Except for sign and a common factor, these are equivalent to the solution of (1, 2, 2). References ========== .. [1] The algorithmic resolution of Diophantine equations, Nigel P. Smart, London Mathematical Society Student Texts 41, Cambridge University Press, Cambridge, 1998. FrrrN)r7rrsr)rrrr%r$r%rZs rDparametrize_ternary_quadraticr sYn*"E:C 444S@A!D #- sOS) ) 4rFc 0SU;deUup4n[U5nUcgURS5S:agUS:Xa&USUSsUS'US'[XCU4Xb5upxn XU 4$Uupn [SSS9upn[ SUR 555n[ UR[XU 4X-X-U-X-U-4555nURU SS 9unnUU-n UU-[ UU - U-5- n UU-[ UU - U-5- n [XU 5$) Nrrfrrzr, p, qTrc3.# UH upX-v M g7frWr)r}rrfs rDr1_parametrize_ternary_quadratic..[ s +]TQQS]s)r) rrrrrr>rrNr]rr)rPrurr$r%rZrfy_px_pz_prSrrrrerreq_1rrs rDrr@ sO E>>MCc T A {!~~aA " axqT1Q4 !ad6 sOQ' #}GA!i.GA! +U[[] + +B BGGC q AE1519aeai023 4D   q  .DAq #A 3!A#a% A 3!A#a% A qQ rFc[USS9up#nU[R:XaJ[X#5n[ U5S:a[ U5SupgnOSupgnU(a[ XgU4X#5$XgU4$g)a] Solves the quadratic ternary diophantine equation, `ax^2 + by^2 + cz^2 = 0`. Explanation =========== Here the coefficients `a`, `b`, and `c` should be non zero. Otherwise the equation will be a quadratic binary or univariate equation. If solvable, returns a tuple `(x, y, z)` that satisfies the given equation. If the equation does not have integer solutions, `(None, None, None)` is returned. Usage ===== ``diop_ternary_quadratic_normal(eq)``: where ``eq`` is an equation of the form `ax^2 + by^2 + cz^2 = 0`. Examples ======== >>> from sympy.abc import x, y, z >>> from sympy.solvers.diophantine.diophantine import diop_ternary_quadratic_normal >>> diop_ternary_quadratic_normal(x**2 + 3*y**2 - z**2) (1, 0, 1) >>> diop_ternary_quadratic_normal(4*x**2 + 5*y**2 - z**2) (1, 0, 2) >>> diop_ternary_quadratic_normal(34*x**2 - 3*y**2 - 301*z**2) (4, 9, 1) FrrrfN)r7r=rurtrHrrrs rDr-r-h s{>*"E:C 5:::,S8 s8a< IaLMCc,MCc 13- -};rFc\^[U4SjT55n[X S9R5$)Nc32># UH oTU-v M g7frWrrjs rDr1_diop_ternary_quadratic_normal.. s )5aq\5rEr)rr=r)rrrs ` rDrtrt s' )5 ) )B ,R B H H JJrFc \[XU5n[SU55n[[XE5VVs/sH upgXgS--PM snn5=nupn [X5n X-n X-n [X5n X-n X-n [X5nX-n X-n X-n X-n X-n U(aXXXU 44$XU 4$s snnf)a Return `a', b', c'`, the coefficients of the square-free normal form of `ax^2 + by^2 + cz^2 = 0`, where `a', b', c'` are pairwise prime. If `steps` is True then also return three tuples: `sq`, `sqf`, and `(a', b', c')` where `sq` contains the square factors of `a`, `b` and `c` after removing the `gcd(a, b, c)`; `sqf` contains the values of `a`, `b` and `c` after removing both the `gcd(a, b, c)` and the square factors. The solutions for `ax^2 + by^2 + cz^2 = 0` can be recovered from the solutions of `a'x^2 + b'y^2 + c'z^2 = 0`. Examples ======== >>> from sympy.solvers.diophantine.diophantine import sqf_normal >>> sqf_normal(2 * 3**2 * 5, 2 * 5 * 11, 2 * 7**2 * 11) (11, 1, 5) >>> sqf_normal(2 * 3**2 * 5, 2 * 5 * 11, 2 * 7**2 * 11, True) ((3, 1, 7), (5, 55, 11), (11, 1, 5)) References ========== .. [1] Legendre's Theorem, Legrange's Descent, https://public.csusm.edu/aitken_html/notes/legendre.pdf See Also ======== reconstruct() c38# UHn[U5v M g7frW)rEr}rRs rDrsqf_normal.. s-A}Qrr)rr?r]r)rrPr~rJABCrMrRrsqfrrrpcpapbs rDrKrK sD aA C -- -Bc#l;lsq1d7l;<>> from sympy.solvers.diophantine.diophantine import square_factor >>> square_factor(24) 2 >>> square_factor(-36*3) 6 >>> square_factor(1) 1 >>> square_factor({3: 2, 2: 1, -1: 1}) # -18 3 See Also ======== sympy.ntheory.factor_.core r)rr\r!r r>)rrSrers rDrErE sF,4 ilA qwwy1ytqTy1 221sA c[[X55nUR5HupEUS:wa [S5eX$-nM U$)a Reconstruct the `z` value of an equivalent solution of `ax^2 + by^2 + cz^2` from the `z` value of a solution of the square-free normal form of the equation, `a'*x^2 + b'*y^2 + c'*z^2`, where `a'`, `b'` and `c'` are square free and `gcd(a', b', c') == 1`. rza and b should be square-free)r!rr>r>)rrrrSrers rDrOrO sE $q*A  6<= =  HrFcUS:XdUS:Xa [S5e[U5[U5:a[X5up#nX$U4$US:XagUS:XagUS:Xag[X5nUcgUS-U- U-nUS:XaUSS4$[ U5Hbn[ [U5U-S5upU (dM&[ U5U-n [X 5upn [U*U -X[--X\-U - X-U-5s $ g) a Return a non-trivial solution to `w^2 = Ax^2 + By^2` using Lagrange's method; return None if there is no such solution. Parameters ========== A : Integer B : Integer non-zero integer Returns ======= (int, int, int) | None : a tuple `(w_0, x_0, y_0)` which is a solution to the above equation. Examples ======== >>> from sympy.solvers.diophantine.diophantine import ldescent >>> ldescent(1, 1) # w^2 = x^2 + y^2 (1, 1, 0) >>> ldescent(4, -7) # w^2 = 4x^2 - 7y^2 (2, -1, 0) This means that `x = -1, y = 0` and `w = 2` is a solution to the equation `w^2 = 4x^2 - 7y^2` >>> ldescent(5, -1) # w^2 = 5x^2 - y^2 (2, 1, -1) References ========== .. [1] The algorithmic resolution of Diophantine equations, Nigel P. Smart, London Mathematical Society Student Texts 41, Cambridge University Press, Cambridge, 1998. .. [2] Cremona, J. E., Rusin, D. (2003). Efficient Solution of Rational Conics. Mathematics of Computation, 72(243), 1417-1441. https://doi.org/10.1090/S0025-5718-02-01480-1 rz!A and B must be non-zero integersrrrrrrrrNr)r>rldescentr'r rrr)rrrrrSrr'rRrrKr{Wrrs rDrr sT Ava<== 1vA1.aQwAvAvBwAy AaAAv"ax a[#CFaK3  6q'!)Cq&GA!r!tacz137AE!G< < rFc[U5[U5:a[X5up#nX$U4$US:XagUS:XagX*:XagX:Xa[SU5up$nX-XB4$[X5n[XPU5upgUS-XS--- U-n[ U5n XS--n [X 5upn [ Xk-X-U --X{-Xl--X-U -5$)a Returns a non-trivial solution, (x, y, z), to `x^2 = Ay^2 + Bz^2` using Lagrange's descent method with lattice-reduction. `A` and `B` are assumed to be valid for such a solution to exist. This is faster than the normal Lagrange's descent algorithm because the Gaussian reduction is used. Examples ======== >>> from sympy.solvers.diophantine.diophantine import descent >>> descent(3, 1) # x**2 = 3*y**2 + z**2 (1, 0, 1) `(x, y, z) = (1, 0, 1)` is a solution to the above equation. >>> descent(41, -113) (-16, -3, 1) References ========== .. [1] Cremona, J. E., Rusin, D. (2003). Efficient Solution of Rational Conics. Mathematics of Computation, 72(243), 1417-1441. https://doi.org/10.1090/S0025-5718-02-01480-1 rrrrrrrr)rrLr'gaussian_reducerEr)rrrSrrrr$rZrt_2t_1x_1z_1y_1s rDrLrL? s8 1vA!-aQwAvAvBwv"a.aQ{AqQ'HC a!F( q A  C Av+CAOMCc sws*CGcg,=sws{ KKrFc^[T5mU4SjnUS4nX -S:aUS4OU*S4nUS-US-T-:aXTpTU"XD5U"XU5=n:a;U"XE5U-nXTSXuS-- USXuS-- 4pTU"XD5U"XU5=n:aM;USUS- USUS- 4nU"X5U"XD5s=::aSU"XE5-::aU$ U$U$)a Returns a reduced solution `(x, z)` to the congruence `X^2 - aZ^2 \equiv 0 \pmod{b}` so that `x^2 + |a|z^2` is as small as possible. Here ``w`` is a solution of the congruence `x^2 \equiv a \pmod{b}`. This function is intended to be used only for ``descent()``. Explanation =========== The Gaussian reduction can find the shortest vector for any norm. So we define the special norm for the vectors `u = (u_1, u_2)` and `v = (v_1, v_2)` as follows. .. math :: u \cdot v := (wu_1 + bu_2)(wv_1 + bv_2) + |a|u_1v_1 Note that, given the mapping `f: (u_1, u_2) \to (wu_1 + bu_2, u_1)`, `f((u_1,u_2))` is the solution to `X^2 - aZ^2 \equiv 0 \pmod{b}`. In other words, finding the shortest vector in this norm will yield a solution with smaller `X^2 + |a|Z^2`. The algorithm starts from basis vectors `(0, 1)` and `(1, 0)` (corresponding to solutions `(b, 0)` and `(w, 1)`, respectively) and finds the shortest vector. The shortest vector does not necessarily correspond to the smallest solution, but since ``descent()`` only wants the smallest possible solution, it is sufficient. Parameters ========== w : int ``w`` s.t. `w^2 \equiv a \pmod{b}` a : int square-free nonzero integer b : int square-free nonzero integer Examples ======== >>> from sympy.solvers.diophantine.diophantine import gaussian_reduce >>> from sympy.ntheory.residue_ntheory import sqrt_mod >>> a, b = 19, 101 >>> gaussian_reduce(sqrt_mod(a, b), a, b) # 1**2 - 19*(-4)**2 = -303 (1, -4) >>> a, b = 11, 14 >>> x, z = gaussian_reduce(sqrt_mod(a, b), a, b) >>> (x**2 - a*z**2) % b == 0 True It does not always return the smallest solution. >>> a, b = 6, 95 >>> min_x, min_z = 1, 4 >>> x, z = gaussian_reduce(sqrt_mod(a, b), a, b) >>> (x**2 - a*z**2) % b == 0 and (min_x**2 - a*min_z**2) % b == 0 True >>> min_x**2 + abs(a)*min_z**2 < x**2 + abs(a)*z**2 True References ========== .. [1] Gaussian lattice Reduction [online]. Available: https://web.archive.org/web/20201021115213/http://home.ie.cuhk.edu.hk/~wkshum/wordpress/?p=404 .. [2] Cremona, J. E., Rusin, D. (2003). Efficient Solution of Rational Conics. Mathematics of Computation, 72(243), 1417-1441. https://doi.org/10.1090/S0025-5718-02-01480-1 c8>USUS-TUS-US--$rnr)rrfrs rD_dotgaussian_reduce.._dot s*tAaDy1QqT6!A$;&&rFrrrrro) rrrPrrrfdvrr~s ` rDrru sF AA' AA#(A!RA !tadQh1 q*d1j( ) J" Q4!aD&=!A$Q4-01 q*d1j( ) 1!adQqTk"A AzT!Z/1T!Z</0 H1HrFc[U5(aSU-nOUS-nX4-U-nSnX0S--XAS--XRS--pn X-U :waUS:Xa [S5eGO{XUpn [XU 5U::aGOe[XmU *5=nunnSU;aGOKUU-U -UU-U --*X^-nn[ UU5n[ U5(a/[ UU5n[UU- 5[R::deOKUU-n[UU-UU--UU--5(aUS- n[UU- 5[R::deUUS--UUS---UUS---nUU-U -UU-U --UU-U--n[ U U-SU-U-- U5n[ U U-SU-U-- U5n[ UU-SU-U-- U5n[SXU455(deUS- nGM[W W W4Vs/sHn[U5PM sn5$s snf)an Simplify the solution `(x, y, z)` of the equation `ax^2 + by^2 = cz^2` with `a, b, c > 0` and `z^2 \geq \mid ab \mid` to a new reduced solution `(x', y', z')` such that `z'^2 \leq \mid ab \mid`. The algorithm is an interpretation of Mordell's reduction as described on page 8 of Cremona and Rusin's paper [1]_ and the work of Mordell in reference [2]_. References ========== .. [1] Cremona, J. E., Rusin, D. (2003). Efficient Solution of Rational Conics. Mathematics of Computation, 72(243), 1417-1441. https://doi.org/10.1090/S0025-5718-02-01480-1 .. [2] Diophantine Equations, L. J. Mordell, page 48. rrzbad starting solutionNrc38# UHoRv M g7frW)rrs rDrholzer.. s3A<<r)rr>maxrr rrrrHalfrrr?rL)rSrrrrPr~rsmallstept1t2t3r$r%rZuvrrfrerrrrrrRs rDrNrN sF( Aww aC qD CEE D !tVQ!tVQ!tV 7b=qy !899 a# rr?e # (#66TQ 2: 1S1Q3s7"#QU1 QN 88q!$Aq1u:' ''1AAaC!A#I!O$$Qq1u:& && q!tVa1f_qAv % qSWqs3w 1S ( SUQqSU]A & SUQqSU]A & SUQqSU]A &3!33333  C F 3S/2/Q#a&/2 332sG8c[USS9up#nU[R:XaBUcSnO[SU[ U54-SS9n[ [U5R US95S$g) a Solves the general pythagorean equation, `a_{1}^2x_{1}^2 + a_{2}^2x_{2}^2 + . . . + a_{n}^2x_{n}^2 - a_{n + 1}^2x_{n + 1}^2 = 0`. Returns a tuple which contains a parametrized solution to the equation, sorted in the same order as the input variables. Usage ===== ``diop_general_pythagorean(eq, param)``: where ``eq`` is a general pythagorean equation which is assumed to be zero and ``param`` is the base parameter used to construct other parameters by subscripting. Examples ======== >>> from sympy.solvers.diophantine.diophantine import diop_general_pythagorean >>> from sympy.abc import a, b, c, d, e >>> diop_general_pythagorean(a**2 + b**2 + c**2 - d**2) (m1**2 + m2**2 - m3**2, 2*m1*m3, 2*m2*m3, m1**2 + m2**2 + m3**2) >>> diop_general_pythagorean(9*a**2 - 4*b**2 + 16*c**2 + 25*d**2 + e**2) (10*m1**2 + 10*m2**2 + 10*m3**2 - 10*m4**2, 15*m1**2 + 15*m2**2 + 15*m3**2 + 15*m4**2, 15*m1*m4, 12*m2*m4, 60*m3*m4) FrNz%s1:%iTrrr)r7rrurrHrr)rrrrr%rs rDr.r. sq2+2U;C &+++ =FXC(994HF&r*00F0CDQGG ,rFc[USS9up#nU[R:Xa![[U5R US95$g)aX Solves the equation `x_{1}^2 + x_{2}^2 + . . . + x_{n}^2 - k = 0`. Returns at most ``limit`` number of solutions. Usage ===== ``general_sum_of_squares(eq, limit)`` : Here ``eq`` is an expression which is assumed to be zero. Also, ``eq`` should be in the form, `x_{1}^2 + x_{2}^2 + . . . + x_{n}^2 - k = 0`. Details ======= When `n = 3` if `k = 4^a(8m + 7)` for some `a, m \in Z` then there will be no solutions. Refer to [1]_ for more details. Examples ======== >>> from sympy.solvers.diophantine.diophantine import diop_general_sum_of_squares >>> from sympy.abc import a, b, c, d, e >>> diop_general_sum_of_squares(a**2 + b**2 + c**2 + d**2 + e**2 - 2345) {(15, 22, 22, 24, 24)} Reference ========= .. [1] Representing an integer as a sum of three squares, [online], Available: https://proofwiki.org/wiki/Integer_as_Sum_of_Three_Squares Frr)N)r7rrurIrrrrrr%s rDr0r01 sID*"E:C ',,,&r*00u0=>>-rFc[USS9up#nU[R:Xa![[U5R US95$g)a[ Solves the equation `x_{1}^e + x_{2}^e + . . . + x_{n}^e - k = 0` where `e` is an even, integer power. Returns at most ``limit`` number of solutions. Usage ===== ``general_sum_of_even_powers(eq, limit)`` : Here ``eq`` is an expression which is assumed to be zero. Also, ``eq`` should be in the form, `x_{1}^e + x_{2}^e + . . . + x_{n}^e - k = 0`. Examples ======== >>> from sympy.solvers.diophantine.diophantine import diop_general_sum_of_even_powers >>> from sympy.abc import a, b >>> diop_general_sum_of_even_powers(a**4 + b**4 - (2**4 + 3**4)) {(2, 3)} See Also ======== power_representation Frr)N)r7rrurIrrs rDr2r2Y sH6*"E:C *///)"-33%3@AA0rFc## U(aUc [X5Hn[U5v M g[SUS-5H6n[X5H$n[U5nSU[U5- -U-v M& M8 g7f)a Returns a generator that can be used to generate partitions of an integer `n`. Explanation =========== A partition of `n` is a set of positive integers which add up to `n`. For example, partitions of 3 are 3, 1 + 2, 1 + 1 + 1. A partition is returned as a tuple. If ``k`` equals None, then all possible partitions are returned irrespective of their size, otherwise only the partitions of size ``k`` are returned. If the ``zero`` parameter is set to True then a suitable number of zeros are added at the end of every partition of size less than ``k``. ``zero`` parameter is considered only if ``k`` is not None. When the partitions are over, the last `next()` call throws the ``StopIteration`` exception, so this function should always be used inside a try - except block. Details ======= ``partition(n, k)``: Here ``n`` is a positive integer and ``k`` is the size of the partition which is also positive integer. Examples ======== >>> from sympy.solvers.diophantine.diophantine import partition >>> f = partition(5) >>> next(f) (1, 1, 1, 1, 1) >>> next(f) (1, 1, 1, 2) >>> g = partition(5, 3) >>> next(g) (1, 1, 3) >>> next(g) (1, 2, 2) >>> g = partition(5, 3, zeros=True) >>> next(g) (0, 0, 5) Nrr)r5r?rJrH)rrrrRrs rD partitionr sm\ AI#A)A(N*q!a%A'-!HAAJ'!++.!sA5A7c[U5nUS-S:wag[U5(d [S5eUS-S:XaSnOEUS-S:XaS nO9US-S ;aSnO-S n[X5S:Xa[ U5n[X5S:XaM[ XS- U5nUnUS-U:aXU-pUS-U:aM[ X!-5[ U54$) a Represent a prime `p` as a unique sum of two squares; this can only be done if the prime is congruent to 1 mod 4. Parameters ========== p : Integer A prime that is congruent to 1 mod 4 Returns ======= (int, int) | None : Pair of positive integers ``(x, y)`` satisfying ``x**2 + y**2 = p``. None if ``p`` is not congruent to 1 mod 4. Raises ====== ValueError If ``p`` is not prime number Examples ======== >>> from sympy.solvers.diophantine.diophantine import prime_as_sum_of_two_squares >>> prime_as_sum_of_two_squares(7) # can't be done >>> prime_as_sum_of_two_squares(5) (1, 2) Reference ========= .. [1] Representing a number as a sum of four squares, [online], Available: https://schorn.ch/lagrange.html See Also ======== sum_of_squares rrNzp should be a prime numberr r)rr)r/r%r>rr#powrL)rerPrs rDprime_as_sum_of_two_squaresr sV q A1uz 1::5661uz  R1  Q&  Qla! AQla AAvqA A Q$(a%1 Q$( JA rFc b0SS_SS_SS_SS_S S _S S _S S_SS_SS_SS_SS_SS_SS_SS_SS_SS _S!S"_S#S$0En[U5nUS%:a [S&5eUS%:Xag'[US(5upSU-nUS)-S*:Xag+X;a [XVs/sHo2U-PM sn5$[ US5upEU(aS%S%X$-4$US)-S:XazUS-(dUS-n[ US,S-5HXnXS-- S-n[ U5(dM[U5up[[X&-X(U --U[X- 5-/55s $ eUS-US-- (dUS-n[ US,S-5HFnXS-- n[ U5(dM[U5up[[X&-X(-X)-/55s $ es snf).a Returns a 3-tuple $(a, b, c)$ such that $a^2 + b^2 + c^2 = n$ and $a, b, c \geq 0$. Returns None if $n = 4^a(8m + 7)$ for some `a, m \in \mathbb{Z}`. See [1]_ for more details. Parameters ========== n : Integer non-negative integer Returns ======= (int, int, int) | None : 3-tuple non-negative integers ``(a, b, c)`` satisfying ``a**2 + b**2 + c**2 = n``. a,b,c are sorted in ascending order. ``None`` if no such ``(a,b,c)``. Raises ====== ValueError If ``n`` is a negative integer Examples ======== >>> from sympy.solvers.diophantine.diophantine import sum_of_three_squares >>> sum_of_three_squares(44542) (18, 37, 207) References ========== .. [1] Representing a number as a sum of three squares, [online], Available: https://schorn.ch/lagrange.html See Also ======== power_representation : ``sum_of_three_squares(n)`` is one of the solutions output by ``power_representation(n, 2, 3, zeros=True)`` rrrrr)rrr )rrr")rrr:)rrrU)rr)rr )rr )r rir)rri)rri)rri)rri)r!ir)rr$i )r 'i%)89rr"n should be a non-negative integer)rrrrrrNrr) r/r>rr?rrJr%rsortedr) rspecialrfrRrrKrSr(rrs rDsum_of_three_squaresr s^b^q)^Q ^1i^Y^I^9^ )^-0*^>A:^ORT]^J^ #Z^14l^DG^[^#' ^7;L^KOP\^G q A1u=>>Av !Q>> from sympy.solvers.diophantine.diophantine import sum_of_four_squares >>> sum_of_four_squares(3456) (8, 8, 32, 48) >>> sum_of_four_squares(1294585930293) (0, 1234, 2161, 1137796) References ========== .. [1] Representing a number as a sum of four squares, [online], Available: https://schorn.ch/lagrange.html See Also ======== power_representation : ``sum_of_four_squares(n)`` is one of the solutions output by ``power_representation(n, 2, 4, zeros=True)`` rr)rrrrrrrrr)rr)r/r>rrr?r)rrfrrSrrs rDsum_of_four_squaresrZsZ q A1u=>>Av !Q>> from sympy.solvers.diophantine.diophantine import power_representation Represent 1729 as a sum of two cubes: >>> f = power_representation(1729, 3, 2) >>> next(f) (9, 10) >>> next(f) (1, 12) If the flag `zeros` is True, the solution may contain tuples with zeros; any such solutions will be generated after the solutions without zeros: >>> list(power_representation(125, 2, 3, zeros=True)) [(5, 6, 8), (3, 4, 10), (0, 5, 10), (0, 2, 11)] For even `p` the `permute_sign` function can be used to get all signed values: >>> from sympy.utilities.iterables import permute_signs >>> list(permute_signs((1, 12))) [(1, 12), (-1, 12), (1, -12), (-1, -12)] All possible signed permutations can also be obtained: >>> from sympy.utilities.iterables import signed_permutations >>> list(signed_permutations((1, 12))) [(1, 12), (-1, 12), (1, -12), (-1, -12), (12, 1), (-12, 1), (12, -1), (-12, -1)] rrc3&# UHo*v M g7frWrrs rDr'power_representation..s?1BsNrz9 Expecting positive integers for `(p, k)`, but got `(z, z)`rrrrrc3,># UH oT-v M g7frWr)r}rRrfs rDrrs/Q!Qsrr)rrrrrrr)rrrrrr))rr) rrrrr%rrrT)r/rr?r>r0r"rrr_can_do_sum_of_squaresrrpow_rep_recursivereversedrJ)rrerrrRrberPrrrfeasiblerrfs @rDrrs\$%),)Qvay),GA!1u q5)1"aE:??**;1uA!%   Av q&LAv 6$J !VJ  q!Ba|4'MAvQ///Av 6!Q)!/ 2v!q&QV:'1']Av1 <<q! QS]@]Av!EE 4 -a0 0 Av!a% 1  "Q%!)q. v AQK +A ."1B2A $ $3 A !! $q!A&qQA6HQqu%56777 U-B 0sJ JC J ?JFJ c## US::dUS::agX!:agU[X5-U:agUS:XaUS:Xa[U5v gUS:Xa*[X$5upVU(aXP::a[X5/-5v gUS:aPUS:aI[SUS-5H5nU[XT5- nUS:a g[ XQS- XsU/-U5ShvN M7 gggN 7frn)rr?rrJr)n_ir n_remainingrre next_termexactresiduals rDrrs ax163s;$Av+"El a*;: Y% +, , !8A"1cAg. &Y)::a<,YAxR[Q\I\^_``` /8 asBC ? C c#:# [USX5ShvN gN7f)aReturn a generator that yields the k-tuples of nonnegative values, the squares of which sum to n. If zeros is False (default) then the solution will not contain zeros. The nonnegative elements of a tuple are sorted. * If k == 1 and n is square, (n,) is returned. * If k == 2 then n can only be written as a sum of squares if every prime in the factorization of n that has the form 4*k + 3 has an even multiplicity. If n is prime then it can only be written as a sum of two squares if it is in the form 4*k + 1. * if k == 3 then n can be written as a sum of squares if it does not have the form 4**m*(8*k + 7). * all integers can be written as the sum of 4 squares. * if k > 4 then n can be partitioned and each partition can be written as a sum of 4 squares; if n is not evenly divisible by 4 then n can be written as a sum of squares only if the an additional partition can be written as sum of squares. For example, if k = 6 then n is partitioned into two parts, the first being written as a sum of 4 squares and the second being written as a sum of 2 squares -- which can only be done if the condition above for k = 2 can be met, so this will automatically reject certain partitions of n. Examples ======== >>> from sympy.solvers.diophantine.diophantine import sum_of_squares >>> list(sum_of_squares(25, 2)) [(3, 4)] >>> list(sum_of_squares(25, 2, True)) [(3, 4), (0, 5)] >>> list(sum_of_squares(25, 4)) [(1, 2, 2, 4)] See Also ======== sympy.utilities.iterables.signed_permutations rN)r)rrrs rDrr>sZ$Aq!333s c(US:agUS:agUS:XagUS:Xa [U5$US:XaKUS;ag[U5(a US-S:Xagg[S[U5R 555$US :Xa[ US5SS -S :g$g) a Return True if n can be written as the sum of k squares, False if it cannot, or 1 if ``k == 2`` and ``n`` is prime (in which case it *can* be written as a sum of two squares). A False is returned only if it cannot be written as ``k``-squares, even if 0s are allowed. rFrTr)rrrc3T# UHupUS-S:g=(d US-S:Hv M g7f)rrrrNr)r}rers rDr)_can_do_sum_of_squares..s,L7Ktq1q519*A *7Ks&(rrr)r$r%rr!r>r)rrs rDrrns 1u1uAvAv|Av ; 1::1uzLy|7I7I7KLLLAva|A"a'' rFrW)T)rrLrPrLrrLrzset[tuple[int, int]])F)rrLrrLrPrLrztuple[int, int]r)NF) __future__rsympy.core.addrsympy.core.assumptionsrsympy.core.containersrsympy.core.exprtoolsrsympy.core.functionrsympy.core.mulr sympy.core.numbersr r sympy.core.intfuncr r rrrsympy.core.relationalrsympy.core.singletonrsympy.core.sortingrrsympy.core.symbolrrsympy.core.sympifyrsympy.external.gmpyrrrr$sympy.functions.elementary.complexesr#sympy.functions.elementary.integersr(sympy.functions.elementary.miscellaneousrsympy.matrices.denserrsympy.ntheory.factor_r r!r"sympy.ntheory.generater#sympy.ntheory.primetestr$r%sympy.ntheory.modularr&sympy.ntheory.residue_ntheoryr'r(sympy.polys.polyerrorsr)sympy.polys.polytoolsr*r+rr,sympy.solvers.solvesetr-sympy.utilitiesr.sympy.utilities.miscr/r0sympy.utilities.iterablesr1r2r3r4r5__all__rIr9rrrrrr7r=r^rzrrrrrr$rur3rrMrrrr6rr&rr7joinrfunc_docr*rr/rr+r r rGrFrHrnrkrIrwr r}r rr,rsrrrrr-rtrKrErOrrLrrNr.r0r2rrrrr sum_of_powersrrr)r6s0rDr2sq"4'-(3II$"8-'==559=DD,63A33,0,3PP / *gSgTK&K&\(@C $CL[-[|*$;*&T(?TnE"9EPJ$;J(F"9F(D1DNK0K\>'>6<4<B !%!  1A A0@*oo0@ A  $* "#t44|~ F: !d3S)B% P)XK 6r34  A=H3LlU p>4B(/sD'A HF%?PBL5,pC LZz@/F|8~% a>-4`cUBsK