p^4 mod 120

In this video the Numberphile people prove in 2 ways that \(p^2 \equiv 1\ (mod\ 24)\). That is that \(p^2 - 1\) is divisible by 24, for \(p\), a prime greater than 3.

The second proof involves considering the 3 numbers \(p-1, p, p+1\). There are three consecutive numbers, so one of them must be divisible by 3 but it can't be \(p\). Also one of \(p-1, p+1\) is divisible by 4 and the other is divisible by 2. That means that \((p-1)(p+1)\) is divisible by \(2{\times}4{\times}3 = 24\). So, expanding out the product, \(p^2 \equiv 1\ (mod\ 24)\)

So then I wondered what if we look at one number either side of these three. We still have the previous factors but now we must also have a 5 and if we now say \(p > 5\) then the 5 can't be in \(p\) so we know that $$(p-2)(p-1)(p+1)(p+2) \equiv 0\ (mod\ 120)$$

Grouping things in +/- pairs we get $$(p^2-1)(p^2-4) \equiv 0\ (mod\ 120)$$

and multiplying out $$p^4 - 5p^2 + 4 \equiv 0\ (mod\ 120)$$ From the earlier result we know that \(p^2-1 \equiv 0\ (mod\ 24)\) so \(5p^2-5 \equiv 0\ (mod\ 120)\). Adding that into the previous result gives us $$p^4 - 1 \equiv 0\ (mod\ 120)$$

So now we have \(p^4 \equiv 1\ (mod\ 120)\) for prime \(p>5\).

Sadly throwing \(p-3\) and \(p+3\) in too doesn't get us a similar result, like \(p^6 \equiv 1\ (mod\ 840)\). With some fiddling around, we can get $$p^6 + 14p^2-14 \equiv 1\ (mod\ 840)$$ but nothing simpler jumps out.