Prove that $(x+y)^p=x^p+y^p \mod p$ where $p$ is a prime. Let $p$ be a prime, we want to prove that $(x+y)^p=x^p+y^p \mod p$.
The left handside looks like something we can throw the binomial theorem at, also in the previous proof the book asked to prove that $p|\binom{p}{k}$ for all natural numbers $k$ except $0$ and $p$. So we then know that $\binom{p}{k} \equiv 0 \mod p$ for k unequal to $0$ and to p$.

$$(x+y)^p=\sum^p_{k=0} \binom{p}{k} x^{p-k} y^k \equiv \binom{p}{p}x^{p-p}y^p+  \binom{p}{0}x^{p-0}y^0 \mod p $$
 Since all other terms are congruent to $0$.
This can be simplified to:

$$(x+y)^p=x^p+y^p\mod p $$

Alternatively, for a way shorter proof we can just note that for any prime $p$ we have that $a^p \equiv a \mod p$ by Fermat's little theorem. $$ (x+y)^p \equiv x+y \mod p \equiv x^p +y^p \mod p$$

Is this correct?
I had made a mistake in the binomial theorem and had made a wrong assumption about $p| \binom{p}{k}$. k cannot be equal to $0$ or to $p$. This fixed the problem I had. The comments were very useful. I was also able to apply Fermat's little theorem and this gave a really nice short proof.
I had made a mistake in the binomial theorem and had made a wrong assumption about $p| \binom{p}{k}$. k cannot be equal to $0$ or to $p$. This fixed the problem I had. The comments were very useful. I was also able to apply Fermat's little theorem and this gave a really nice short proof.