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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!

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2 Answers 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?

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$0\mod (a+1)\implies a(a-1)0 = 0$? –  Tomi Sebastián Juárez Dec 6 '13 at 19:47
    
Just right! Not too much, not too little! +1 –  amWhy Dec 8 '13 at 14:33

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

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