Let´s take any x>3$x>3$ and choose a,b such that a<b$a<b$,a-b$a-b$ even and ab=x^2-1$ab=x^2-1$. Then with y=(b-a)/2$y=(b-a)/2$, z=(b+a)/2$z=(b+a)/2$ we have x^2 + y^2 = z^2 + 1$x^2 + y^2 = z^2 + 1$. Particular cases∶ if x$x$ even then x^2 + ((x^2-2)/2)^2 = (x^2/2)^2 + 1;$x^2 + ((x^2-2)/2)^2 = (x^2/2)^2 + 1$; if x$x$ odd then x^2 + ((x^2-5)/4)^2 = ((x^2+3)/4)^2 + 1$x^2 + ((x^2-5)/4)^2 = ((x^2+3)/4)^2 + 1$.
