Why does $(x+y)^p=x^p+y^p$ over $\mathbb{Z_2}$?

Question

My first thought was to expand it with the binomial theorem and then see that all the coefficients are $0$ mod 2 except the first and the second, but this is not true, unless i'm missing something. Like if p=13 then the coefficient of the second term in the expantion would be 13 which reduces to 1 mod 2.

Answer 1

Just take values $x,y\in{\Bbb Z}_2$ and check the equation. If $x=y=0$, then $(0+0)^p=0=0^p+0^p$. If either $x=1$ or $y=1$, then $(1+0)^p =1^p= 1^p+0^p$. If $x=y=1$, then $(1+1)^p = 0^p=0$ and $1^p+1^p=0$.

Answer 2

Over $\mathbb{Z}_2$, if $x+y \equiv 0$, then $x \equiv y$ and we have $x^p \equiv y^p$ and $x^p + y^p \equiv 0$.

If $x+y \equiv 1$, WLOG if $x \equiv 0$, then $y \equiv 1$, so $x^p + y^p \equiv 1$.

Answer 3

In $\mathbb{Z}_2$, the identity $x^p = x$ holds. So $(x+y)^p = x+y = x^p + y^p$.