Challenge Level

Wiktor from LAE Tottenham in the UK, Mahdi from Mahatma Gandhi International School in India, Nikita from Jersey College for Girls in Jersey and David sent in similar answers. This is Wiktor's work:

Nikita, David, Wiktor and Mahdi all began in the same way. This is Nikita's work:

Mahdi's proof is very similar, but David proved the statement without writing out the cases algebraically. David wrote:

$n^5-n^3 = n^3(n+1)(n-1)$

If $n$ is even, $8|n^3$

If $n$ is odd, then one of $(n-1), (n+1)$ is divisibly by $4$ and the other is divisible by $2$. So $(n-1)n^3(n+1)$ is necessarily divisible by $8.$

Nikita, David and Mahdi's work all begins in the same way. This is Mahdi's work:

David considered five cases without writing them out algebraically:

$n^5-n=(n-1)n(n+1)(n^2+1)$

If $n\equiv 0 \text{ (mod 5})$ then $5|n$

If $n\equiv 1 \text{ (mod 5})$ then $5|n-1$

If $n\equiv \pm 2 \text{ (mod 5})$ then $5|n^2+1$ (which means $n\equiv 2 \text{ (mod 5})$ or $n\equiv 3 \text{ (mod 5)}$

If $n\equiv 4 \text{ (mod 5})$ then $5|n+1$

Nikita used the same idea but on the unfactorised expression:

Wiktor used proof by induction:

Wiktor, Mahdi and Sahin sent in very similar proofs. This is Sahin's work:

Nikita and Wiktor sent in a different proof. This is Nikita's work:

(for all *positive *integers $n$)

David sent in a different proof, using modular arithmetic. My maths teacher would have said it's like using a sledgehammer to crack an egg: