Consider the following theorem:
If $(x,y,z)$ are the lengths of a Primitive Pythagorean triangle, then $$x = r^2-s^2$$ $$y = 2rs$$ $$z = r^2+z^2$$ where $\gcd(r,s) = 1$ and $r,s$ are of opposite parity.
According to the previous theorem,My try is the following:
since $x = r^2-s^2$, $x$ is difference of two squares implying that $x \equiv 0 \pmod 4$. But $x=21 \not \equiv 0 \pmod 4$. Hence, there are no triangles having such $x$.
Is that right?
My argument is false here. Please refer to the appropriate answer.