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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.

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  • $\begingroup$ Here's a MathJax tutorial :) $\endgroup$ – Shaun Jun 6 '14 at 19:23
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    $\begingroup$ Fixed the expression fonts. :) $\endgroup$ – Wal Jun 6 '14 at 19:47
  • $\begingroup$ math.stackexchange.com/questions/816681/… In this subject the decision drew. What is the problem? $\endgroup$ – individ Jun 7 '14 at 4:44
  • $\begingroup$ 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? $\endgroup$ – Wal Jun 7 '14 at 5:20
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    $\begingroup$ It's from An Introduction to Diophantine Equations by Titu Andreescu and Dorin Andrica (2002), pg 79. $\endgroup$ – Wal Jun 9 '14 at 1:25
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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 $k=1,m=2,n=-1.$

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  • $\begingroup$ Interesting idea, coffeemath. The modulus makes sense after that, but is that a 'legal' move (replacing $n$ with $-n$)? $\endgroup$ – Wal Jun 8 '14 at 14:14
  • $\begingroup$ @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. $\endgroup$ – coffeemath Jun 8 '14 at 20:15
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    $\begingroup$ @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. $\endgroup$ – coffeemath Sep 3 '16 at 1:23
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    $\begingroup$ 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. $\endgroup$ – Will Jagy Sep 3 '16 at 1:36
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    $\begingroup$ @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). $\endgroup$ – coffeemath Sep 3 '16 at 2:02
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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
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  • $\begingroup$ $${{\left( ps+kms+hk\right) }^{2}}+{{\left( hp-ks\right) }^{2}}+m\,\left( hp-ks\right) \,\left( ps+kms+hk\right) =\left( {{p}^{2}}+kmp+{{k}^{2}}\right) \,\left( {{s}^{2}}+hms+{{h}^{2}}\right) $$Maybe because $$(p^2+kmp+k^2)(s^2+hms+h^2)=z^2$$ not only if $$p=h, k=s$$. $\endgroup$ – AlexSam Sep 3 '16 at 16:37

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