Right triangle with rational sides and area = 1 equivalent to n = 3 case of Fermat's Last Theorem I watched a talk by Andrew Wiles in which he spoke about the proof of Fermat's Last Theorem, and he said something that has puzzled me. He mentioned that the $n=3$ case of FLT (i.e. proving $a^3+b^3=c^3$, where $abc\neq0$, has no integer solutions) was equivalent to proving that there cannot be a right triangle with rational sides and area $=1$.
I have been trying to deduce why that is the case. I haven't gotten very far with it, but my thoughts so far are as follows:
Assume there is a right triangle with rational sides and area $=1$. The side-lengths would have a least common denominator, call it $d$. Then we can say the legs are $\frac{m}{d}$, $\frac{n}{d}$, and the hypotenuse is $\frac{p}{d}$, for some positive integers $m, n$, and $p$.
Pythagoras gives: $\frac{m^2}{d^2}+\frac{n^2}{d^2}=\frac{p^2}{d^2}$, or just $m^2+n^2=p^2$.
The area condition gives $\frac12\cdot\frac{m}{d}\cdot\frac{n}{d}=1$, or equivalently $mn=2d^2$.
This is my first dead end. Try as I might to do manipulations, I can't see how the above implies the existence of integers $a, b, c$ satisfying $a^3+b^3=c^3$.
So I thought I'd try in reverse, and start by assuming some non-zero integers satisfy $a^3+b^3=c^3$.
One thought is to factor the sum of cubes on the LHS, to get $(a+b)(a^2-ab+b^2)=c^3$.
Then, thinking of Pythagoras, I can rearrange to isolate $a^2+b^2$, which looks like $a^2+b^2=\frac{c^3}{a+b}+ab$.
However here is where I am stuck again. I don't really have a good idea of what to do with the right side.
I'd appreciate help connecting the dots. Just FYI, I don't know much at all about elliptic curves, other than their name and that they were something Wiles used in his full proof of FLT. So I'm hoping there are  elementary approaches to finish my work. But I'd also like to know if there is no elementary way to finish what I started, and elliptic curves (or something else) are necessary.
 A: Could you please provide a link to the actual talk you saw so others can check you heard things correctly?  I ask this because the nonexistence of a rational-sided right triangle with area 1 is well-known in number theory to be equivalent to the equation $x^4 + y^4 = z^2$ having no solution in nonzero (equivalently, positive) integers $x$, $y$, and $z$. That equation is close to, but not the same as, the Fermat equation of degree $4$ (which defines a projective curve of genus 3). I suspect that what Wiles had in mind is the connection between rational-sided right triangles of area $1$ and FLT for $n = 4$, not $n=3$.
FLT for $n=3$ is equivalent to the equation $y^2 = x^3 + 16$ having no solution in nonzero rational numbers, as shown in the last paragraph here. The nonexistence of a rational-sided right triangle of area $1$ is equivalent to the equation $y^2 = x^3 - x$ having no solution in nonzero rational numbers $x$ and $y$. The (elliptic) curves $y^2 = x^3 - x$ and $y^2 = x^3 + 16$ are not the same thing (and are not isogenous).
Update: In Weil's "Number Theory: An Approach Through History from Hammurapi to Legendre," Weil discusses the Fermat equation of degree 3 on pp. 114-117, saying Fermat claimed to prove it by descent but never revealed what he did, and then Weil discusses Euler's work on this equation. Euler is generally credited with the first solution of FLT for degree $3$. (His approach had a gap involving subtleties about $\mathbf Z[\sqrt{-3}]$, which was later filled in.) On pp. 76-79 Weil discusses letters of Fermat and Frenicle about there not being a rational-sided right triangle with area 1 or 2 (equivalently, area equal to a square or twice a square) and he points out the connection of these to nonsolvability of $x^4 \pm y^4 = z^2$ in positive integers. In algebraic number theory books, the Fermat equation of degree 4 is always discussed in connection with $x^4 + y^4 = z^2$. That equation is never brought up when discussing the Fermat equation of degree $3$. Fermat is credited with showing there is no rational-sided right triangle with area 1 and with solving FLT in degree 4 because the latter follows from the former, but he is not credited with solving FLT in degree $3$ despite claiming he could do so since we have no details of such a proof by him. In short, I think Wiles made a mistake by linking right triangles with area $1$ to FLT in degree $3$.
A: It is enough to consider integer Pythagorean triangles, enough to consider only primitive triangles, and say the are cannot also be a square. Let's see, there are coprime integers $x,y$    giving sides $$ 2xy,  x^2 - y^2  $$
and area half that product. So there is no (nontrivial) integer solution to
$$ xy(x^2 - y^2)  = z^2 $$
I also don't know how the quartic provides an elliptic curve
