Where is the flaw that led to the fallacy that $x^4 + y^4 = z^4$?

Question

Consider $x^2 + y^2 = r^2$. Then take the square of this to give $(x^2 + y^2)^2 = r^4$. Clearly, from this $r^4 \neq x^4 + y^4$.

But consider: let $x=a^2, y = b^2 $and$\,\,r = c^2$. Sub this into the first eqn to get $(a^2)^2 + (b^2)^2 = (c^2)^2$. $x = a^2 => a = |x|,$ and similarly for $b.$

Now put this in to give $|x|^4 + |y|^4 = r^4 => (-x)^4 + (-y)^4 = r^4 $ or $ (x)^4 + (y)^4 = r^4,$ both of which give $ x^4 + y^4 = r^4$ Where is the flaw in this argument?

Many thanks.

Answer 1

$x = a^2$ does not imply that $a = |x|$, rather $|a| = \sqrt{x}$.

Answer 2

The fact that $x=a^2$ is quite far to imply that $a=|x|$ (second paragraph).

Answer 3

Note that $(x^2+y^2)^2=r^4$ does not imply that $r^4\ne x^4+y^4$.

In fact, you show that $$a^4+b^4=c^4$$ provided $x^2+y^2=r^2, x=a^2, y=b^2, z=c^2$. So what?