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?
 A: Expanding on previous comment: $\,(x+1)^p-x^p-1\,$ is divisible by $\,x^2+x+1\,$ as polynomials.
Let $\,\omega_1,\omega_2\,$ be the roots of $x^2+x+1\,$, and note that they are the two complex cube roots of unity since $x^3-1=(x-1)(x^2+x+1)$. By Vieta's relations $1+\omega_1+\omega_2=0\,$. It also follows that $1+\omega_1^p+\omega_2^p=0$ for any $p \ge 1$ which is not a multiple of $3\,$.
Then,  $\,P(\omega_1)=(\omega_1+1)^p-\omega_1^p-1=(-\omega_2)^p-\omega_1^p-1=-\omega_2^p-\omega_1^p-1=0\,$ for any odd $p \gt 3$ which is not a multiple of $3\,$. The same goes to show that $\,P(\omega_2)=0\,$, so $\,P(x)\,$ is a polynomial multiple of $\,(x-\omega_1)(x-\omega_2)=x^2+x+1\,$.
Let $\,(x+1)^p-x^p-1=Q(x)(x^2+x+1)\,$ where $\,Q(x)\,$ is a polynomial of degree at most $p-2$ with integer coefficients. Then:
$$\left(\frac{x}{y}+1\right)^p-\frac{x^p}{y^p}-1=Q\left(\frac{x}{y}\right)\left(\frac{x^2}{y^2}+\frac{x}{y}+1\right)$$
Multiplying by $\,y^p\,$ gives:
$$(x+y)^p-x^p-y^p=y^{p-2}Q\left(\frac{x}{y}\right) \cdot \left(x^2+xy+y^2\right)$$
Quite obviously, $\,y^{p-2}Q\left(\cfrac{x}{y}\right)\,$ is an integer whenever $x,y$ are integers.
As a note, the above only used the conditions that $p > 3$ must be an odd integer not divisible by $\,3\,$ (though not necessarily a prime).
A: 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 $4$.
