Congruence equation proof

Question

Proof that $\forall{a}\in\Bbb N \rightarrow a^3\equiv a\mod (a+1)$

I do not know how to prove these equations.

I only know that $a\equiv m \mod b \implies m | ( b- a ) \implies b-a=m\times k $ for some $k \in \Bbb Z$

So (...)

$(a+1)-a^3 \implies (-a^3+a)+a(-a^2+1)+1=a\times k$

But I can not take it from here.

I wish comeone could help me to solve this.

Thanks!

Answer 1

$$a^3-a=a(a^2-1)=a(a+1)(a-1)\equiv0\pmod{a+1}$$

Answer 2

We have (Not as you said) $$a\equiv m \mod b\iff b|(a-m)$$ and since $$a^3-a=a(a^2-1)=a(a-1)(a+b)$$ can you take it from here?