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Fermat challenged Frenicle with finding a pythagorean triple (a,b,c) where $(a-b)^2-2b^2$ is itself a square. By making the substitution $a=m^2-n^2$, $b=2mn$, and $c=m^2+n^2$ into $(a-b)^2-2b^2=d^2$ we obtain the following quartic:

$m^4-4m^3n-6m^2n^2+4mn^3+n^4=d^2$

The goal of the exercise is to somehow obtain a curve that corresponds to this equation, and to then use another curve, namely a quadratic intersecting it with this curve to generate another solution from the solution (1,0,1).

This method was used to show that there also exists a pythagorean triple (a,b,c) where c and a-b are both squares. Using the curve $y^2=2x^4-1$ and using an intersecting quadratic to generate a nontrivial solution.

Any hints or tips would be appreciated.

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  • $\begingroup$ Do you know if $a,b,c$ are required to be natural numbers? If not, then $(-45,108,117)$ works and I can explain how I got it. Otherwise, I will keep looking. $\endgroup$ – Álvaro Lozano-Robledo Feb 26 '14 at 3:27
  • $\begingroup$ Ah! Got it, $(a,b,c)=(6068, 624, 6100)$. $\endgroup$ – Álvaro Lozano-Robledo Feb 26 '14 at 3:32
  • $\begingroup$ May I ask for the source of this problem? Where did you find it? $\endgroup$ – Álvaro Lozano-Robledo Feb 26 '14 at 3:51
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Let us begin with the quartic $$m^4-4m^3n-6m^2n^2+4mn^3+n^4=d^2.$$ Dividing through by $n^4$, we obtain $$m'^4-4m'^3-6m'^2+4m'+1=d'^2,$$ with $m'=m/n$ and $d'=d/n^2$. Now we can homogenize this equation to obtain $$E:M^4-4M^3N-6M^2N^2+4MN^3+N^4=D^2N^2,$$ which has a rational point $P=[M,N,D]=[0,0,1]$. The curve $E$ is in fact non-singular, and of genus $1$, so the pair $(E,P)$ is an elliptic curve. An appropriate change of variables (left to the OP as she/he asks for hints) brings $E$ to a Weierstrass model $$E':y^2 - 4xy - 32y = x^3 + 20x^2 + 96x,$$ and there is a rational map $\phi:E'\to E$. The Mordell-Weil group of $E'$ is isomorphic to $\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}$, generated by $S=[-8,0,1]$, and $T=[-12,-16,1]$. Now, there is an appropriate $\mathbb{Z}$-linear combination of $S$ and $T$ (again left to the OP to figure out) that gives a point $R$, such that $\phi(R)=[78/1343,4/1343,1]$. This says that $[M,N,D]=[78/1343,4/1343,1]$ is a point on $E$, and therefore $(M,N,DN)$ is a solution of the original quartic. This would give a pythagorean triple with the desired quality, but with rational coefficients, so instead we may realize that $$(1343M,1343N,1343^2DN)=(78,4,4\cdot 1343)$$ is also a point in the original quartic. This point corresponds with $$(a,b,c)=(6068, 624, 6100)$$ and indeed $(a-b)^2-2b^2=28858384=2^4\cdot 17^2\cdot 79^2$ is a square.

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