# Quadratic Diophantine Equation $x^2 + axy + y^2 = z^2$

$x^2 + axy + y^2 = z^2$

where x, y, z are integers to be solved and a is a given integer.

All integral solutions are given by

$x = k(an^2 - 2mn), y = k(m^2 - n^2), z = k(amn - m^2 - n^2)$ and

$x = k(m^2 - n^2), y = k(an^2 - 2mn), z = k(amn - m^2 - n^2)$

(due to diagonal symmetry in x and y)

where $m,n$ are integers with $\gcd(m,n) = 1,$ but $k \in \mathbb Q$ is rational such that $(a^2 - 4) \, k \in \mathbb Z.$ This is Theorem 2.3.2. on page 90 of An Introduction to Diophantine Equations by Andreescu, Andrica, and Cucurezeranu. (2010). EDIT BY WILL JAGY.

I have no problem understanding how the solution forms were derived; they were just basic algebraic manipulation. But then when it comes to the solutions in positive integers, the form becomes

$x = k(an^2 + 2mn), y = k(m^2 - n^2), z = k|amn + m^2 + n^2|$ and

$x = k(m^2 - n^2), y = k(an^2 + 2mn), z = k|amn + m^2 + n^2|$

where k, m, n are positive integers, an + 2 m > 0 and m > n.

What I can understand is that we apply modulus to the x, y, and z in the previous form to get the latter form (we want x, y, and z to be in positive integers), but I can't seem to understand how an + 2 m > 0 and m > n work to prove

$|x| = |k(an^2 - 2mn)| = kn|an - 2m| = kn(an + 2m) = k(an^2 + 2mn)$ and

$|z| = |k(amn - m^2 - n^2)| = k|amn + m^2 + n^2|$.

Can anyone help me on this? I've been pondering for almost a week. It's driving me crazy. Thank you in advance.

• Jun 6, 2014 at 19:23
• Fixed the expression fonts. :)
– Wal
Jun 6, 2014 at 19:47
• math.stackexchange.com/questions/816681/… In this subject the decision drew. What is the problem? Jun 7, 2014 at 4:44
• I can't seem to understand how $an + 2m > 0$ and $m > n$ work to prove $|x|=|k(an2−2mn)|=kn|an−2m|=kn(an+2m)=k(an2+2mn)$ and $|z|=|k(amn−m2−n2)|=k|amn+m2+n2|$ for solutions in positive integers. How do I explain it?
– Wal
Jun 7, 2014 at 5:20
• It's from An Introduction to Diophantine Equations by Titu Andreescu and Dorin Andrica (2002), pg 79.
– Wal
Jun 9, 2014 at 1:25

There is something wrong here. The given set of formulas does give infinitely many solutions, that part is fine.

Example: $$x^2 + 8 xy + y^2 = z^2$$ Set of solutions that does not fit those formulas: $$x = 16 u^2 - 14 u v + 3 v^2, \; \; y = -2 u^2 + 2 u v, \; \; z = 2 u^2 + 6 u v - 3 v^2$$ I know this is new because the indefinite binary form $2 u^2 + 6 u v - 3 v^2$ is neither the principal form nor its negative: it does not represent $+1$ or $-1$ over the integers.

I have requested two books by Andreescu through my city library, a local college has them. I cannot imagine that they claim all solutions come up through one set of formulas.

pari
? x =  16 * u^2 - 14 * u * v + 3 * v^2
%1 = 16*u^2 - 14*v*u + 3*v^2
? y = -2 * u^2 + 2 * u * v
%2 = -2*u^2 + 2*v*u
? z = 2 * u^2 + 6 * u * v - 3 * v^2
%3 = 2*u^2 + 6*v*u - 3*v^2
? x^2 + 8 * x * y + y^2 - z^2
%4 = 0

==========================================================

jagy@phobeusjunior:~$./Conway_Positive_Primes 1 8 1 1000 5 1 8 1 original form 1 6 -6 Lagrange-Gauss reduced Represented (positive) primes up to 1000 61 109 181 229 241 349 409 421 541 601 661 709 769 829 ========================================================== jagy@phobeusjunior:~$ ./Conway_Positive_Primes 2 6 -3   1000 5
2           6          -3   original form

2           6          -3   Lagrange-Gauss reduced

Represented (positive) primes up to  1000

2     5    17    53   113   137   173   197   233   257
293   317   353   557   593   617   653   677   773   797
857   953   977

• $${{\left( ps+kms+hk\right) }^{2}}+{{\left( hp-ks\right) }^{2}}+m\,\left( hp-ks\right) \,\left( ps+kms+hk\right) =\left( {{p}^{2}}+kmp+{{k}^{2}}\right) \,\left( {{s}^{2}}+hms+{{h}^{2}}\right)$$Maybe because $$(p^2+kmp+k^2)(s^2+hms+h^2)=z^2$$ not only if $$p=h, k=s$$. Sep 3, 2016 at 16:37

If in the general integer solution, the parameter $n$ is replaced by $-n$, then one has $$\{x,y\}=\{k(an^2+2mn),\ k(m^2-n^2)\},\ z=k(-amn-m^2-n^2).$$ This still describes all integer solutions, provided $m,n,k$ are allowed to be any integers. In an attempt to get only the solutions for which $x,y,z>0$ it might be tempting to suppose $m,n,k$ to be positive and then impose $m>n$ and also $an+2m>0,$ in order to force $x,y>0$, and then also place absolute values around the $z$ formula, to make $z>0$ also (since perhaps $a<0$ that would be necessary).

However doing this one will miss some positive solutions. Consider the case of $a=5$ and the solution $(x,y,z)=(1,3,5)$ to the equation $x^2+5xy+y^2=z^2.$ If one insists on positive $m,n$ as in the above formulas, then it must be that $k=1$ (since gcd of 1,3,5 is 1), and also $m^2-n^2=1$ is not possible as then $m=1,n=0$ against positivity. So it must be that $m^2-n^2=3,$ so that $m=2,n=1$ But the values $(m,n,k)=(2,1,1)$ lead to the wrong value for the other variable. That is, in the "positive $m,n,k$" formulation, if it is $x$ which is $3$, then $y$ doesn't come out $1$ under $(m,n,k)=(2,1,1).$ Instead it comes out $1\cdot(5\cdot 1^2+2 \cdot 2 \cdot 1)=9.$

Note that if we allow negative $m$ or $n$ we can get the solution $x,y,z=1,3,5$ using $k=1,m=2,n=-1.$

• Interesting idea, coffeemath. The modulus makes sense after that, but is that a 'legal' move (replacing $n$ with $-n$)?
– Wal
Jun 8, 2014 at 14:14
• @Wal If one says "$x=2n,y=-n$" where $n$ can be any integer, positive negative or zero, that's the same as saying "$x=2(-n)=-2n,y=-(-n)=n$" where $n$ can be any integer, positive negative or zero. In other words for describing a solution using letters allowed to be any integer, the same solution is described if one takes one of the letters and replaces it by its negative in all the formulas. Jun 8, 2014 at 20:15
• @WillJagy I see. My answer was only based on IF the OP description gave all the solutions, in terms of integers $k,m,n,$ THEN nothing changes when say $n$ gets replaced by $-n.$ Your answer below shows a single parmetrization is not generally enough. Sep 3, 2016 at 1:23
• Right, I thought you were just going along with the OP. This came to my attention at recent math.stackexchange.com/questions/1909192/… where they are, in essence, doing the same thing: if we parametrize all rational solutions and then multiply through by the denominator, we don't get all solutions, although we do get (unpredictable) integer multiples of all solutions. Sep 3, 2016 at 1:36
• @WillJagy I noticed the same thing when I tried to pasrametrize something like $x^2+ky^2=z^2$ by dividing by $z^2$ first, then parametrizing rational solutions, then remultiplying. It didn't give all the integer solutions to the original, and I had to resort to a set of Pell equations (I forget details). Sep 3, 2016 at 2:02