4
$\begingroup$

Given 2 primitive pythagorean triples with parameters as per: https://en.wikipedia.org/wiki/Pythagorean_triple#Generating_a_triple

First triple has parameters $(m,n)$. Second has parameters $(r,s)$. Note that $m>n$ and $r>s$. We define $g$ and $h$ according to these equations:

$$g^2 = (mn(r^2-s^2))^2 + (mn(r^2-s^2)-rs(m^2-n^2)+2mnrs)^2$$ $$h^2 = (rs(m^2-n^2))^2 + (mn(r^2-s^2)-rs(m^2-n^2)+2mnrs)^2$$

I'm trying to show that $g$ and $h$ can't both be integers unless $m=r$ and $n=s$.

If you're interested where this comes from, I was working on the rational distance problem: http://unsolvedproblems.org/index_files/RationalDistance.htm

And it got down to proving the above problem. Appreciate any tips on solving the above problem.

$\endgroup$
  • $\begingroup$ What makes you so sure that the conjecture is even true? $\endgroup$ – J.Gudal Dec 12 '15 at 9:47
  • $\begingroup$ I'm not sure. I did a limited computer search and couldn't find solutions. I could find solutions to either equation but not both simultaneously. $\endgroup$ – Ameet Sharma Dec 12 '15 at 9:50
  • $\begingroup$ This system has no solutions. That's for sure. There is one option if to find a point on the continuation of one of the ribs. This changes solve the system of equations. $\endgroup$ – individ Dec 12 '15 at 10:16
  • $\begingroup$ individ, how do you know there are no solutions? $\endgroup$ – Ameet Sharma Dec 12 '15 at 10:19
1
+100
$\begingroup$

This is NOT a complete answer, since I believe that proving this would be extraordinarily difficult. Finding a possible counterexample is also highly non-trivial.

We can simplify the problem by defining $e=m/n$ and $f=r/s$, so we deal with rational variables. Both equations are of the form "quartic in $e$ equals a square". The first equation becomes

\begin{equation*} \Box=f^2e^4-2f(f^2+2f-1)e^3+2(f^4+2f^3-f^2-2f+1)e^2+2f(f^2+2f-1)e+f^2 \end{equation*}

Since $e=0$ gives a solution, this quartic is birationally equivalent to an elliptic curve. Using standard methods, we find the curve

\begin{equation*} V^2=U(U-2f^2(f+1)^2)(U-2(f^2+1)(f^2+2f-1)) \end{equation*} with the reverse transformation \begin{equation*} e=\frac{2f^6+8f^5+8f^4-f^2(U+2)-2f\,U+U-V}{2f(f^4+4f^3+4f^2-U-1)} \end{equation*}

The curves, for fixed $f$, have rank at least $1$, with a point of infinite order at $(\,4f^2(f^2+1), 4f^2(f-1)^2(f^2+1)\,)$ assuming $|f| \ne 0,1$.

Thus one possible method to find a solution is to select $f$, find the generators of the elliptic curve, get points on the curve and find $e$. Then test whether $e$ and $f$ satisfy the second equation.

The second equation also gives a quartic in $e$, namely \begin{equation*} \Box=2f^2e^4-2f(f^2+2f-1)e^3+(f^4+4f^3-2f^2-4f+1)e^2+2f(f^2+2f-1)e+2f^2 \end{equation*} which can also be shown to be equivalent to a different elliptic curve, since $e=1$ gives a solution. This elliptic curve also has rank at least $1$ for fixed $f$.

I hope this small contribution helps in some way.

$\endgroup$
  • $\begingroup$ Dear Allan, thanks for your contribution. Is there a text you recommend for studying elliptic curves and birational equivalence? Seems this is related to a lot of problems I'm interested in. $\endgroup$ – Ameet Sharma Dec 16 '15 at 10:52
  • 1
    $\begingroup$ "Rational Points on Elliptic Curves" by Joseph Silverman and John Tate is an excellent introduction. $\endgroup$ – Allan MacLeod Dec 16 '15 at 11:21
  • $\begingroup$ One thing I noticed is that the properties of m,n as parameters of pythagorean triples aren't used here. Is there any way to make use of the fact that m>n, r>s... So e>1 and f>1. Also m,n are relatively prime. So are r,s. So e and f are not integers. Do these facts somehow help? $\endgroup$ – Ameet Sharma Dec 19 '15 at 14:15
  • $\begingroup$ @AllanMacLeod: If we let $$e = \frac{f^2-2f-1}{4f}$$ then the first eqn becomes a square. Substituting this into the second eqn yields the condition, $$2(5+12f+10f^2-12f^3+5f^4) = y^2$$ though I don't know if this can have a solution. $\endgroup$ – Tito Piezas III Dec 29 '15 at 16:02
  • $\begingroup$ According to John Cremona's ratpoint program, this quartic is not locally soluble at the prime 2, so no solution exists. $\endgroup$ – Allan MacLeod Dec 29 '15 at 16:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.