I need to analyze if $(a+b)^{m} \equiv a^{m}+b^{m} \pmod{m}$ for $m=2$ and $m=4$
For both values of $m$ i worked whith the definition of congruence.
For $m=2$ works out and then $(a+b)^{m} \equiv a^{m}+b^{m} \pmod{m}$
For $m=4$ i have that
$4$ divides $(a+b)^{4} \equiv a^{4}+b^{4} \pmod{4}$ $\Rightarrow$ $4$ divides $4a^{3}b + 6a^{2}b^{2} + 4ab^{3}$ $\Rightarrow$
$(a+b)^{m} \equiv a^{m}+b^{m} \pmod{m}$ $\Leftrightarrow$ $a^{2}b^{2}$ multiple of $2$.
I don't know if what i did is ok. Any other way to analyze this? Maybe a property of congruence?