<s>[INST] Prove that $x^p+y^p \equiv (x+y)^p \pmod{x^2+xy+y^2}$ 
Let $p > 3$ be a prime, and let $x,y$ be integers such that $\gcd(x,y) = 1$. Prove that $$x^p+y^p \equiv (x+y)^p \pmod{x^2+xy+y^2}.$$

I thought about expanding $(x+y)^p$ using the binomial theorem, but I didn't see how that would help because the modulus is an algebraic expression. How can we prove the statement?
 [/INST] Not true, you can use polynomial modular arithmetic, i will give an example for what i am talking about :
when $p=3$ we have $(x+y)^3=x^3+3x^2 y+3x y^2+y^3 $, and we want to find the remainder module $x^2+x y+y^2$ , the leading expression in the bigger polynomial is $x^3$ which means we need to multiply the smaller polynomial by $x$ and subtract from each, we arrive at $ 2 x^2 y+2 x y^2+y^3$ and the leading $x$ term is $2x^2 y$ in this new polynomial, so we need to multiply by $2y$ and subtract, we arrive at $-y^3$ which is the smallest term we can reach.
Now for the expression $(x+y)^p = ? \mod (x^2+x y+y^2)$ we have $6$ cases :
$p=6n$ => $y^{6n}$
$p=6n+1$ => $xy^{6n}+y^{6n+1}$
$p=6n+2$ => $xy^{6n+1}$
$p=6n+3$ => $-y^{6n+3}$
$p=6n+4$ => $-xy^{6n+3}-y^{6n+4}$
$p=6n+5$ => $-xy^{6n+4}$
And for the expression $x^p+y^p=? \mod (x^2+xy+y^2)$ we have $3$ cases :
$p=3n$ => $2y^{3n}$
$p=3n+1$ => $xy^{3n}+y^{3n+1}$
$p=3n+2$ => $-xy^{3n+1}$.
And because $5,5+6,5+2*6,5+3*6,5+4*6$ are all primes, one can conclude mistakenly that  it works just for primes, but actually it also works for $p=5+6*5=35$, also $p=77$.
In General : it works for any $p=\pm 1 \mod 6$.
For Instance : take $x=2,y=3,p=35$ , so $2^35+3^35 = 5^35 \mod 19$, checking it on Wolfram Alpha, they both give remainder $4$.
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