<s>[INST] Show $5n^3+7n^5 \equiv 0$ mod 12 for all integers $n$. I've been stuck on this question for quite a while now. I could obvious show this by setting $n$ to all numbers from 0-11, however this is not very efficient. 
Note $5n^3+7n^5 = n^3(5+7n^2)$. We could use the Chinese remainder theorem, we show $n^3(5+7n^2) \equiv 0 $ mod 3 and mod 4, but this doesn't seem to be any easier. What would be the best way to solve this problem?
 [/INST] Oh, geez.....  $7+5 = 12$ so $5n^3 + 7n^5 \equiv 5n^3 - 5n^5 \equiv -7n^3 + 7n^5$.
Suffices to show $12|n^5 - n^3 = n^3(n^2 - 1)= n^3(n+1)(n-1)$.
.....
$n-1, n, n+1$ are three consecutive integers so $3$ divides one of them.
$n-1,n,n+1$ are two consecutive integers so either $2|n$ in which case $8|n^3$, or $2|n-1$ and $2|n+1$ in which case $4|(n-1)(n+1)$.  In either case $4|n^3(n+1)(n-1)$.
So $12|n^3(n+1)(n-1)$  so
$n^3(n+1)(n-1) \equiv 0 \mod 12$
$n^5 - n^3 \equiv 0 \mod 12$
$7n^5 - 7n^3 \equiv 0 \mod 12$
$7n^5 + 5n^3 \equiv 0 \mod 12$.
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