Germain Primes and Solutions of $x^p+ y^p+ z^p =0$ I am taking a second course on number theory which is primarily of algebraic number theory. So, in the course , professor gave an exercise related to Germain primes and it's application to Fermat's Last Theorem. I think I will need help on this.
Question : Let $p\geq 3$ be a prime and 2p+1 also be prime , then prove that for every (x,y,z) which is solution of $x^p+y^p+z^p=0$,  the congruence $xyz\equiv 0$(mod p) holds.
( Note that my question is not following :Let p be a Germain prime with q = 2p + 1, and suppose that $x^p + y^p = z^p$ (mod q). Then at least one of x, y, z is divisible by q. This result was found by me on a website of a university when I googled my question.) I also know that atleast I didn't knew where to use the fact that 2p+1 is also prime with mu question statement. But , I have checked statement in notes atleast 3 times and so prove or supply a contradiction using my statement of the question only.
Attempt: I have to show that  p divides xyz. But p is prime so p must divide either one of x or y or z atleast.  Now , I thought of using Fermat's Little theorem to equation $x^p+y^p+z^p=0$ but I am not sure that either (x,p)=1 or (y,p)=1 or (z,p)=1. I also can see that $x^p+ y^p +z^p\equiv 0 $( mod p) but again I need a hint or two on how to move foreward.
Kindly help me!
 A: Only the sketch of a proof (found on the internet on https://www.math.uni-kiel.de/ in German):
W.l.o.g. we assume that $x,y,z$ are pairwise coprime. Then we consider
\begin{align*}
–x^p &=\underbrace{ (y + z)}_{=:a_1}\underbrace{ (y^{p–1} – y^{p–2}z + y^{p–3}z^2 – ⋅⋅⋅ + z^{p–1})}_{=:a_2}\\
–y^p &=\underbrace{ (z + x)}_{=:b_1} \underbrace{(z^{p–1} – z^{p–2}x + z^{p–3}x^2 – ⋅⋅⋅ + x^{p–1})}_{=:b_2}\\
–z^p &= \underbrace{(x + y)}_{=:c_1}\underbrace{ (x^{p–1} – x^{p–2}y + x^{p–3}y^2 – ⋅⋅⋅ + y^{p–1})}_{=:c_2}
\end{align*}
Next, if one of the pairs $(a_1,a_2),(b_1,b_2),(c_1,c_2)$ has a common divisor, then it can be shown that it is $\pm p$ and thus that $p\mid xyz.$
If otherwise all pairs $(a_1,a_2),(b_1,b_2),(c_1,c_2)$ are coprime, then all numbers $a_1,a_2,b_1,b_2,c_1,c_2$ are powers of $p.$ Then we have to deduce a contradiction by using Fermat's little theorem, which is used to prove:
Let $p$ be a Sophie-Germain prime and $a,x,y,z\in \mathbb{Z}\backslash \{0\}.$ Then $a^{p}=\pm 1 \pmod{(2p+1)}$ or $a^p=0\pmod{2p+1}.$ Thus $x^p+y^p+z^p=0$ implies $2p+1\mid xyz.$
I admit that there is still significant work to do, but it is at least a road map.
