 Two is not a congruent number

A primitive Pythagorean triple of area twice a square would satisfy $mn(m-n)(m+n)=2s^2$. Pairwise coprimality makes the odd factors squares and the even factor twice a square. The possibility $m$ even is excluded modulo eight. Thus $m=a^2,n=2b^2,m+n=c^2,m-n=d^2$. The coprime integers $u=(c+d)/2,v=(c-d)/2$ have $uv=b^2$, hence are themselves squares, and $a^2=u^2+v^2$ contradicts fourth-power sum cannot be a square. Clearing denominators gives the assertion for rational triangles.