<s>[INST] $n^3 \equiv n^5 \pmod{12} $? I am proving that 
$$5n^3 + 7n^5 \equiv 0 \pmod{12}$$
It would suffice to show 
$$n^3 \equiv n^5 \pmod{12}$$
How would I go about doing that?
I suppose I could just go through each $n \equiv r \pmod{12}$ with $r$ from $1$ to $11$ and show that $n^3 \equiv n^5 \pmod{12}$ for each, but that would be tedious.  Surely there's a better way.
 [/INST] $$\begin{align}n^5-n^3&=(n-2)(n-1)n(n+1)(n+2)-4n^3+4n\\\\
&=(n-2)(n-1)n(n+1)(n+2)-4(n-1)n(n+1)\end{align}$$
The first expression is a product of 5 consecutive numbers, hence is divisible by $5!$ i.e., by 120. Similarly, the second expression is divisible by $4\cdot (3!)$ i.e., by 24.
Also, $n^5-n^3 = n^2(n-1)n(n+1)$. Clearly, a product of 3 consecutive numbers $(n-1)n(n+1)$ is divisible by 3.
Again, $n^5-n^3 = n^3(n^2-1)$. 
If $n$ is even, the R.H.S. is divisible by 8.
If $n$ is odd (equals to $2m+1$, say), then $$n^2-1 = (2m+1)^2 - 1 = 8m(m+1)/2$$ i.e, is divisible by 8 (See If $n$ is an odd natural number, then $8$ divides $n^{2}-1$).
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