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.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

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