Prove that $10 | (n^a - n^b)$.

Question

$n$ is a positive integer. Prove that there exists positive integers $a$ and $b$, $(a > b)$ such that $10 | (n^a - n^b)$.

I have tried to prove this by induction on $n$, but I get stuck at the induction step trying to prove it for $n = k + 1$, not knowing how to expand $((k + 1)^a - (k + 1)^b)$. Is this the wrong approach to solving this problem, or am I missing something here?

Answer 1

As suggested in the comment by Akiva Weinberger, take $a=5$ and $b=1$.

Then $(k+1)^a-(k+1)^b=(k+1)^5-(k+1)=(k+1)[(k+1)^4-1]$

$=(k+1)(k^4+4k^3+6k^2+4k)=(k+1)k(k^3+4k^2+6k+4)$.

This is divisible by $2$ because $(k+1)k$ is,

and modulo $5$ this is the same as $(k+1)k(k-1)(k^2+1)$, so it is divisible by $5$ too,

and therefore it is divisible by $10$.

Answer 2

We have $n^{a} - n^{b} \equiv 0\pmod 2$ since if $n \equiv 1 \pmod 2$ , then $ n^{a} - n^{b}\equiv 1-1 \equiv 0\pmod 2$ and if $n\equiv 0 \pmod 2, n^{a} - n^{b}\equiv0-0 \equiv 0\pmod 2$.

If $5|n$, then $n^{a}-n^{b}\equiv0-0\equiv 0 \pmod 5$ and so $10|n^{a}-n^{b}$.

If $5 \nmid n$, then $n^{4} \equiv 1 \pmod 5$ by Fermat's little theorem so that $n^{8} \equiv 1 \pmod 5 \implies$

$n^{8}-n^{4} \equiv 0 \pmod 5$ and altogether, in this case, $10|n^{a}-n^{b}$ as well where $a=8$ and $b=4$.