Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 19 / 1

There is a necessary ordering qualification: the printed formula requires the even leg to be called $b$. The primitive Pythagorean triple $(a,b,c)=(4,3,5)$ satisfies the printed hypotheses but cannot have $b=2nm$. Thus the intended assertion is true after interchanging the two legs if necessary. We prove that version, including the parity condition on the parameters.

A primitive Pythagorean triple has pairwise coprime sides: a prime dividing any two also divides the third. The two legs cannot both be odd, since their squares would sum to $2(\mathrm{mod}4)$, and cannot both be even. Call the odd leg $a$ and the even leg $b$; then $c$ is odd. The positive integerssatisfy $uv=(b/2{)}^2$ and are coprime, since a common divisor divides both $c$ and $a$. A product of coprime positive integers is a square only when both factors are squares, by their prime factorizations. Write $u=n^2$, $v=m^2$. Then $n>m>0$, $\gcd (n,m)=1$, andFurthermore $n,m$ have opposite parity, since $c$ is odd. Conversely these coprimality and parity conditions give a primitive triple: the square identity follows by expansion, and a common odd prime divisor of its legs would divide both $n,m$, while the odd leg excludes a common factor two. This is the primitive Pythagorean parametrization.

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 isThese 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. WriteBoth $w,z$ are odd. Odd squares are $1(\mathrm{mod}8)$, so $2m=w^2-z^2$ is divisible by eight; hence $m$ is divisible by four and $v$ is even. Define positive integersThey satisfyThus $(r,s,u)$ 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. No integer right triangle has a nonzero square area. This is the Fermat right triangle theorem, with the actual infinite descent construction exhibited.