<s>[INST] Proof by Induction Problem I need to prove that for natural numbers $a$, and positive integers $n$, the number $a^{2n+1}-a$ is divisible by $6$. 
I have proved the case when $n=1$, that $a^3-a$ is divisible by $6$. I'm having trouble proving that if $a^{2n+1}-a$ is divisible by $6$, then $a^{2(n+1)+1}-a$ is also divisible by $6$. 
 [/INST] hint: $a^{2n+1} - a = a(a^{2n}-1) = a(a^2-1)(a^{2n-2} + a^{2n-4} + \cdots + 1) = a(a-1)(a+1)(.....)$ is divisible by $6$ since it has a product of $3$ consecutive integers: $a-1,a,a+1$.
To do an induction proof. 
The base case is done. Assume $6\mid a^{2n+1} - a \Rightarrow a^{2n+3} - a = (a^{2n+3} - a^{2n+1}) + (a^{2n+1} - a) = a^{2n}\cdot a\cdot (a-1)(a+1) + (a^{2n+1}-a)$ is divisible by $6$ since the first term has again a product of $3$ consecutive integers, and the expression in the parentheses is divisible by $6$ by inductive step so the sum is divisible by $6$, complete the induction proof.
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