About three conjectures concerning the Pythagorean Theorem.

The accepted answer covers the first conjecture.

The second and third conjectures are in fact related. To solve,

$$\big(x^2 + (x + 1)^2\big)^2 = y^2+z^2$$

$$a^2+b^2 = (b+1)^2$$

and turns out to have the solution,

$$\big(x^2 + (x + 1)^2\big)^2 = (2x + 1)^2+(\color{brown}{2x^2 + 2x})^2 =(\color{brown}{2x^2 + 2x}+1)^2$$

Suppose that is $c=4k+3$. Then $c|a^2+b^2$ and since it is prime $c|a$ and $c|b$. Thus $a=mc$ and $b=nc$. But then we get $m^2+n^2=1$ which is impossible if $mn\ne 0$.