Since \(y^2=2uv\), exactly one of either \(u\) or \(v\) is even.
Proof:
Assume, for sake of contradiction, both \(u\) and \(v\) are even.
This means \(u=2k\), \(v=2l\), for \(k,l \in \mathbb{Z}\).
It follows \(2|(2k = u)\) and \(2|(2l = v)\). Hence \(u\) and \(v\) share a common factor, 2, so gcd(u,v) \(\geq\) 2.
However, this is a contradiction since \(u\) and \(v\) are coprime, i.e., \(gcd(u,v) = 1\). Thus, our original assumption is wrong.
But, since at least one of \(u\) and \(v\) must be even, but not both, exactly one must be.
Q.E.D.