algebra precalculus - Proof that $sqrt{x}=-sqrt{x}$

$\sqrt{x}=\sqrt{1\cdot x}=\sqrt{(-1)^2\cdot x} = \sqrt{(-1)^2} \cdot \sqrt{x} = (-1) \cdot \sqrt{x}=-\sqrt{x}$

The idea popped into my head while I was evaluating an integral. I have a feeling that I made some obvious mistake because the "proof" is so simple, but I don't see any flaw. Of course, there must be a flaw somewhere. What is it?

Answer

Most fake proofs involving square roots rest, at some point, on the false identity
$$\sqrt{a^2} = a$$
This identity seems natural and true, which is why it fools us. The correct identity is
$$\sqrt{a^2} = |a|$$