2021. okt. 19. Who does not know Diophantus does not deserve the name of math- ... [10] Bernard Frenicle de Bessy Traité des triangles rectangles en ...
We provide explicit formulae for primitive, integral solutions to the Diophantine equa- tionx2+y2=M, whereMis a product of powers of Pythagorean primes, i.e., of primes of the form 4n+1. It turns out that this is a nice application of the theory of
Gaussian integers.
1 Introduction
The history of the Diophantine equationx2+y2=Mhas its roots in the study of Pythagorean triples. The oldest known source is Plimpton 322, a Babylonian clay tablet from around 1800 BC: This table lists two of the three numbers of Pythagorean triples, i.e., integersx;y;zwhich satisfyx2+y2=z2. Euclid's formulaa=m2n2,y= 2mn, z=m2+n2, wheremandnare coprime and not both odd, generates all primitive Pythagorean triples, i.e., triples wherex;y;zare coprime. In 1625 Albert Girard, a French-born mathematician working in Leiden, The Netherlands, who coined the abbreviations sin;cos, and tan for the trigonometric functions and who was one of the rst to use brackets in formulas, stated that every prime of the form 4n+1 is the sum of two squares (see [ 24 ]). Pierre de Fermat [ 19 , tome premier, p.293, tome troisieme, p.243{246] claimed that each suchPythagorean primeand its square is the sums of two squares in a single way, its cube and biquadratic in two ways, its fth and sixth powers in three ways, and so on. It is easy to see that, if an odd prime is a sum of two squares, it 1 must be of the form 4n+ 1. The reverse implication, called Fermat's Theorem on sums of two squares, or Girard's Theorem, is much more dicult to prove. However, Fermat stated in a letter to Carcavi, communicated to Huygens (August 14, 1659, see [ 19 , tome deuxieme, p. 432]) that he had a proof by the method of innite descent for the fact that each Pythagorean prime is the sum of two squares, but he gave no details. Recall that by the Dirichlet Prime Number Theorem (see [ 11 ]), there are innitely many Pythagorean primes. Bernard Frenicle de Bessy who lived 1604{1674 was an advocate of experimental math- ematics: By hisMethode des exclusionshe concluded from looking at numerical tables that, ifp1;p2;:::are distinct Pythagorean primes, then the numberN=pk11pk22pknnis the hypotenuse of exactly 2 n1primitive right triangles (see [9, p. 22{34,156{163]). The theory was nally put on a solid footing by Leonhard Euler who proved Girard's Theorem in two papers (see [ 14 ] and [ 13 ]). In the sequel, 1775, Joseph-Louis Lagrange gave a proof based on his general theory of integral quadratic forms (see [ 21 , p. 351]). The theory of quadratic forms came to a full understanding with Gauss'Disquisitiones arithmeticae[16]. Gauss showed that for odd integersM >2 of the formM=PQ, wherePandQare products of powers of primes of the form 4n+1 and 4n+3, respectively, the Diophantine equationx2+y2=Mis solvable in positive integers if and only ifQis a perfect square (see
Gauss [
17 , p.149f]). Richard Dedekind contributed two more proofs for Girard's Theorem: see [ 10 ,x27, p. 240] and [12, Supplement XI, Ueber die Theorie der ganzen algebraischen Zahlen, p. 444]. Another beautiful proof uses Minkowski's Theorem on convex sets and lattices (see, e.g., [ 25 ,x7.2]). The shortest argument is Don Zagier's famous one-sentence proof [ 27 ] of Girard's Theorem. For a Pythagorean primep, Gauss provided an explicit formula for the unique primitive solutionfx;ygof the Diophantine equationx2+y2=p. Namely, with z:=D12 2n n E we have fx;yg=fz;jhz(2n)!ijg; wherehui 2(p2 ;p2 ) denotes the residue ofumodp(see [8, Chapter 5] for a proof). Another explicit formula was found by Jacobsthal in his dissertation [ 20 ]: The odd number infx;ygis given by12 p X n=1 xp x21p where ( ap ) denotes the Legendre symbol. Both formulae are of more theoretical interest. For an ecient algorithm to compute the primitive solution we refer to [ 26 The purpose of this paper is to provide explicit formulae for primitive, integral solutions to the Diophantine equationx2+y2=M, whereMis a product of powers of Pythagorean primes. 2
2 Combining solutions
A recurring phenomenon in the theory of Diophantine equations is that solutions may be combined to generate new solutions of a given equation. For the equation a
2+b2=M;(1)
this is shown in Lemma 1 . To keep the notation short we write (a;b)Mfor an interger solution of ( 1 ). Trivially, we have (a;b)M=)(b;a)Mand (a;b)M=)(a;b)M. Now, for two pairs of integers (a;b) and (c;d), we dene (a;b)(c;d) := (acbd;ad+bc):(2)
The following result is similar to [
18 , Lemma4]. Lemma 1.Leta;b;~a;~bbe integers and letM;Nbe positive integers such that(a;b)M and(~a;~b)N. Then(a;b)(~a;~b) MN; in other words, we have (a~ab~b;a~b+b~a)MN: Proof.We have to verify that (a~ab~b)2+ (a~b+b~a)2=MN. Indeed, we have (a~ab~b)2+ (a~b+b~a)2= (a2+b2)|{z} =M(~a2+~b2)|{z} =N=MN: q.e.d.
The operation (
2 ) reminds of the product of complex numbers. Indeed, as we shall see below, the Gaussian integersZ[i] are the adequate language to discuss equation (1).
3 Primitive solutions forM=pk
The formulae of Gauss and Jacobsthal yield explicit primitive solutions of ( 1 ) ifMis a Pythagorean primep. Now we want to see how solutions forM=pk,ka positive integer, can be generated from this.
As mentioned above, the product (
2 ) from Section 2 corresp ondsto the complex m ultipli- cation if we consider the rst and second entry as real and imaginary part, respectively.
In particular, Lemma
1 can b eform ulatedas follo ws: Fact 2.Leta;b;~a;~bbe integers and letM;Nbe positive integers such that(a;b)Mand (~a;~b)N. Then, forz:= (a+ib)(~a+i~b), we have
Re(z);Im(z)
MN: 3 So, from now on we will work with Gaussian integersZ[i] =fa+ib:a;b2Zg(see, e.g., [ 15 ] as a general reference): Gaussian integers are a factorial ring, i.e., each element inZ[i] has a unique factorisation up to the units1;i. Every Pythagorean primepcan be decomposed by two Gaussian primes, which are the complex conjugate of each other, i.e., Pythagorean primes are of the formp=for some2Z[i], and this represents the corresponding unique primitive solution of ( 1 ). As an example, 5 can be factorised by
1+2i;12i. This is also true for 2 = (1+i)(1i). On the other hand, all non-Pythagorean
primes inZ, dierent from 2, are also primes inZ[i].
Proposition 3.Letp=be a Pythagorean prime and letkbe a positive integer. ThenjRe(k)j;jIm(k)jis the primitive solution tox2+y2=pk.
Proof.By observing thatpk=k
k, we see that the above equation is satised by jRe(k)j;jIm(k)j. Thus, it only remains to show that these numbers are relatively prime. Assume not, then there exist integersu;v;where >1 such thatk=(u+iv). By the uniqueness of prime factorisation inZ[i] we get=lfor some positive integerl. In particular, arg()
2Qwhich is a contradiction to Niven's Theorem.q.e.d.
Although the formula in Proposition
3 is practically trivial in the con textof Gaussian integers, it does not seem to be very widely known. Indeed, the formulas we now have at hand are missing for the corresponding sequences in theOn-Line Encyclopedia of Integer SequencesOEIS. A few examples: Letp=be a factorised Pythagorean prime,ak= jRe(k)jandbk=jIm(k)j. Then we have: p= 5:xk= minfak;bkgandyk= maxfak;bkgforM= 5kare explicit formulas for the integer sequences [ 1
A230710 ] and [2,A230711 ], respectively.
p= 13:xk= minfak;bkgandyk= maxfak;bkgforM= 13kare explicit formulas for the integer sequences [ 22
A188948 ], and [23,A188949 ], respectively.
p= 17:xk= minfak;bkgandyk= maxfak;bkgforM= 17kare explicit formulas for the integer sequences [ 3
A230622 ], and [4,A230623 ], respectively.
p= 73:xk= minfak;bkgandyk= maxfak;bkgforM= 73kare explicit formulas for the integer sequences [ 5
A230962 ] and [6,A230963 ], respectively.
4 Primitive solutions forM=Qn
l=1pkll In this section we show how one can nd the primitive solution to the Diophantine equation x
2+y2=M, whereMis a product of powers of Pythagorean primes. Also strongly related
with the following part is [ 7 , Lemma3.30].
Theorem 4.Letnandklbe positive integers,pl=l
lbe pairwise distinct Pythagorean primes for1lnand letM=Qn l=1pkll. Then Re nY l=1 kll ;Im nY l=1 kll 4 is a primitive solution forx2+y2=M.
Proof.Obviously, we haveM=Qn
l=1kllQ n l=1kll. Therefore,x2+y2=Mis clearly satised byReQn l=1kll ,ImQn l=1kll It remains to show that our solution is relatively prime. If not, then there exists integers u;v;where >1 such thatQn l=1kll=(u+iv). In this case we must have=Qn l=1k0ll with 0k0lkl. Additionally, it holds true==Qn l=1 lk0l. Observe that all prime factors ofare dierent from1i. Thus, we have a contradiction to the unique prime factorisation inZ[i].q.e.d. The following proposition was stated by Frenicle without a proof, as we mentioned in the introduction. Proposition 5.Letp1;:::;pn,k1;:::;kn, andMbe as in Theorem4 . Then there are 2 n1primitive solutions tox2+y2=M. Proof.LetI,I0be a partition of the setf1;2;:::;ngand M=nY l=1p kll= nY l=1 kll nY l=1 kll Y l2I kllY l2I0kll |{z} =:I Y l2I kllY l2I0kll |{z} I be factorised inZ[i]. Then eachIgives us a primitive solution ofM= Re(I)2+Im(I)2. Conversely, iffx;ygis a primitive solution to the equationx2+y2=M, thenM= (x+iy)(xiy). So, both of these factors can be factorised by the Gaussian primes ofM multiplied by a unit ofZ[i]. Since these factorisations must be the complex conjugates of each other and (x;y) = 1, there existsI f1;2;:::;ngandk2 f0;1;2;3gsuch that x+iy=ikI. This shows that each primitive solution to the equation above can be constructed by the right choice ofI. It remains to show thatx2+y2=Mhas exactly 2n1solutions. For this letI1andI2be subsets off1;2;:::;ngand assume thatI1andI2represent the same solution, i.e., we
Then we nd;0inRsuch that
arg(I1) =+0and arg(I2) =0 and it must hold either +00(mod2 ) or+0 (0) (mod2 5 which implies0 mod4 or00 mod4 . However, sinceand0are either argu- ments of primitive solutions for equations of the formx2+y2=M0, whereM0dividesM, orI1;I2must be disjoint or equal, we conclude the latter. Furthermore, ifI1andI2are disjoint or equal, then () is clearly satised, so we get the same primitive solution. Thus, there are exactly 2 n1dierent choices forIsuch that the resulting primitive solutions are dierent from each other.q.e.d.
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