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.