<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] *

*$n^2 = 0,1,4$ or $9 \mod 12$ by checking each number $0,1,2,3,4,5,6$ (we don't have to check $7,8,9,10$ and $11$ since they are negatives and $x^2 = (-x)^2).$

*$n^4 = (n^2)^2 = n^2 \mod 12$ by checking each number 0,1,4 and 9.

*thus $n^3 = n^5.$

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