Chinese remainder theorem/Fermat's little theorem problem

Question

Prove that $p^4 \equiv 1 \pmod {240}$ for any prime $p>5$.

I'm not sure how to go about this at all - I started with some computation to check it works for $\{7,11,13\}$ and they all end in $1$ - I suspect this has something to do with it. Also, I've tried splitting up $240$ into $60$ and $4$ but I get stuck. $Help^2$

Answer 1

$240=2^4\cdot3\cdot5$ and so $\lambda(240)=\mathrm{lcm}(\frac12\phi(2^4),\phi(3),\phi(5))=\mathrm{lcm}(4,2,4)=4$.

According to Carmichael's theorem, $x^4 \equiv 1 \bmod 240$ if $\gcd(x,240)=1$. This applies to all primes $p > 5$.

Here, $\lambda$ is Carmichael's function, which gives the exponent of the group $U(n)$.

Answer 2

Hint: You can prove the stronger theorem that $2^4 \cdot 3 \cdot 5 \mid n^4-1$ for any integer $n$ that is not a multiple of $2$ or $3$ or $5$. The prime powers have no common factor, so it suffices to prove divisibility by each of them separately. For $3$ and $5$, you can either use Fermat's little theorem or you can just check all the cases modulo $3$ or $5$ respectively. For $2^4$, factorize $n^4-1$ and count how many $2$s appear in the factors.

[Edit: Someone funny decides to downvote a perfectly correct non-circular answer after it has been untouched for half a year!]