 Solution

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.