Show that $(a+b)^n ≡ a^n + b^n (mod 2)$ for all $a,b$ and for $n≥1$

Question

Show that $(a+b)^n ≡ a^n + b^n (mod 2)$ for all $a,b$ and for $n≥1$

What is the solution of this problem? Thanks for the answers!

Answer 1

$$(0+0)^n \equiv 0^n \equiv 0 \equiv 0 + 0 \equiv 0^n + 0^n$$ $$(0+1)^n \equiv 1^n \equiv 1 \equiv 0 + 1 \equiv 0^n + 1^n$$ $$(1+0)^n \equiv 1^n \equiv 1 \equiv 1 + 0 \equiv 1^n + 0^n$$ $$(1+1)^n \equiv 0^n \equiv 0 \equiv 1 + 1 \equiv 1^n + 1^n$$

Answer 2

Lil' Fermat says for any $a$, $a^2\equiv a\mod 2$. An easy induction shows that, as a consequence, $a^n\equiv a\mod 2$ for all $n\ge 1$. Thus the congruence $(a+b)^n\equiv a^n+b^n\;$ becomes simply $$a+b\equiv a+b,$$ which I leave as an exercise.