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I have been reading about this quadratic Diophantine equation of the form

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

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.

Will Jagy
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Wal
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  • Here's a MathJax tutorial :) – Shaun Jun 06 '14 at 19:23
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    Fixed the expression fonts. :) – Wal Jun 06 '14 at 19:47
  • http://math.stackexchange.com/questions/816681/find-all-integers-satisfying-m2-n-12n-1n-2n-22/816685#816685 In this subject the decision drew. What is the problem? – individ Jun 07 '14 at 04:44
  • 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 07 '14 at 05:20
  • Maybe the idea is to change sign on one or more of $m,n,k$ to make the $x,y,z$ come out positive. That is, perhaps the formulation for positive solutions is not using the "same" $m,n,k$ as were used for the description of all integer solutions. – coffeemath Jun 08 '14 at 03:38
  • Look at the formula and select those which always give a positive decision. Or you can rewrite the formula is slightly different, specifying that one number was always bigger than the other. For example so: $s=p+s$ – individ Jun 08 '14 at 04:29
  • I think the reformulation above makes one not get all the positive solutions. (see my "answer" for details, too long to put in a comment.) – coffeemath Jun 08 '14 at 04:54
  • You say you "have been reading" about $x^2+axy+y^2=z^2$. I would like to know where you are reading about it, is it only in previous stackexchange, or in a text, or at a reliable on-line website? It's also not clear whether the formulation for getting positive x,y,z is meant to be something you found done somewhere, or was something you just tried to come up with yourself. – coffeemath Jun 08 '14 at 20:19
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    It's from An Introduction to Diophantine Equations by Titu Andreescu and Dorin Andrica (2002), pg 79. – Wal Jun 09 '14 at 01:25

2 Answers2

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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
Will Jagy
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  • $${{\left( ps+kms+hk\right) }^{2}}+{{\left( hp-ks\right) }^{2}}+m,\left( hp-ks\right) ,\left( ps+kms+hk\right) =\left( {{p}^{2}}+kmp+{{k}^{2}}\right) ,\left( {{s}^{2}}+hms+{{h}^{2}}\right) $$Maybe because $$(p^2+kmp+k^2)(s^2+hms+h^2)=z^2$$ not only if $$p=h, k=s$$. – AlexSam Sep 03 '16 at 16:37
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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 $k=1,m=2,n=-1.$

coffeemath
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  • Interesting idea, coffeemath. The modulus makes sense after that, but is that a 'legal' move (replacing $n$ with $-n$)? – Wal Jun 08 '14 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. – coffeemath Jun 08 '14 at 20:15
  • I see, thanks. :) – Wal Jun 09 '14 at 01:26
  • coffee, it is not true that all solutions come from the given formulas. I am requesting two book by Andreescu that should show the sort of claim that really is made. The theorem is that a finite set of such parametrizations does give all solutions, but not always just one. I give an example in a new answer to this question. – Will Jagy Sep 03 '16 at 01:09
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    @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. – coffeemath Sep 03 '16 at 01:23
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    Right, I thought you were just going along with the OP. This came to my attention at recent http://math.stackexchange.com/questions/1909192/quadratic-diophantine-equation-with-rational-coefficients 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. – Will Jagy Sep 03 '16 at 01:36
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    @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). – coffeemath Sep 03 '16 at 02:02
  • coffee, got one of the Andreescu Andrica books that had a correct statement, I edited that into the question. The whole point, really, is that the letter $k$ there is allowed rational, just with restrictions on the denominator. – Will Jagy Sep 07 '16 at 22:22