= Solution
Suppose an integer <right triangle> with square area exists, and choose one with the least positive hypotenuse $c$. It is primitive. Indeed, dividing all sides by their common divisor $d$ gives an integer right triangle whose area is the square of the rational number $s/d$, where $s^2$ was the original area. Its area is an integer, since one of its legs is even. A rational number whose square is an integer is itself an integer, as follows by writing it in lowest terms. Thus division by $d>1$ would give another square-area triangle with smaller hypotenuse.
By the <primitive Pythagorean parametrization>, after ordering the legs the triangle has sides $n^2-m^2,2nm,n^2+m^2$, with coprime $n>m>0$ of opposite parity. Its area is
$$
nm(n-m)(n+m).
$$
These four positive factors are pairwise <coprime>: the pair $n-m,n+m$ has gcd dividing $2m$ and is odd, and the other pairs have gcd one by $\gcd(n,m)=1$. Since the product is a square, each factor is a square. Write
$$
n=u^2,\qquad m=v^2,\qquad n+m=w^2,\qquad n-m=z^2.
$$
Both $w,z$ are odd. Odd squares are $1\pmod8$, so $2m=w^2-z^2$ is divisible by eight; hence $m$ is divisible by four and $v$ is even. Define positive integers
$$
r=\frac{w+z}{2},\qquad s=\frac{w-z}{2}.
$$
They satisfy
$$
r^2+s^2=\frac{w^2+z^2}{2}=n=u^2,\qquad
\frac{rs}{2}=\frac{w^2-z^2}{8}=\frac m4=\left(\frac v2\right)^2.
$$
Thus $(r,s,u)$ is another positive integer right triangle with square area, but its hypotenuse $u=\sqrt n$ is strictly smaller than $n^2+m^2=c$. This contradicts minimality. \b[No integer right triangle has a nonzero square area.] This is the <Fermat right triangle theorem>, with the actual <infinite descent> construction exhibited.
Back to article page
