How do $x^{p}+y^{p}=z^{p}$ and $x\equiv y \pmod{p}$ together imply $x\equiv -z \pmod{p}$?

Question

I am reading Marcus' Number Fields and I have been a little stuck following his argument in page $4$ (where he is sketching an argument for Case 1 of Fermat's Last Theorem for primes $p$ for which $\mathbb{Z}[\omega]$ is a UFD, where $\omega=e^{2\pi i/p}$). Here is the screenshot:

I don't understand how he deduces $x\equiv -z \pmod{p}$. It seems to be some basic congruence trick, but it is somehow evading me. I can use Fermat's Little Theorem to get $x+y\equiv z\pmod{p}$ from $x^{p}+y^{p}= z^{p} \pmod{p}$. Since $x\equiv y\pmod{p}$, this implies $2x=-z\pmod{p}$. But this is different from what Marcus claims, namely $x\equiv -z \pmod{p}$.

Could someone shed some light on my miserable confusion?

Answer 1

Did you see exercises 16-28 for details?

I believe (might be wrong) the claim there is that if $a^p+b^p=c^p$, then we must have $a\equiv b \pmod p$. He applied that to $a=x, b=y, c=z$ first, and then to $a=x,b=−z,c=−y$.

Answer 2

The book seems to be incorrect. I fully agree with you're reasoning. Additionally, if we assume $x\cong -z$ Then it follows that $y\cong 2x\Rightarrow z\cong 3x$. But $z\cong -x$ so $4x\cong 0$ so $p|x$. This contradicts the assumption in the middle of the paragraph. Additionally, if you apply this to his chain equality we get $GCD(x,y)\geq p$, despite the fact that at the beginning we can say WLOG $GCD(x,y)=1$