<s>[INST] divisibility of $n^{15} - n^3$ by $32760$ I have a question & I have no idea where to begin. I hope someone here can help me. Been stuck for a while.
Prove or disprove: $n^{15} - n^3$ is divisible by $32760$ for all $n \ge 0$.
 [/INST] $32760=2^3\cdot3^2\cdot5\cdot7\cdot13$
Now, $n^{15}-n^3=n^3(n^{12}-1)$
Using Fermat's Little Theorem prime $p\mid(n^p-n)$
So, $5\mid(n^5-n)$ which divides $n^3(n^{12}-1)=n^3(n^4-1)(n^8+n^4+1)$
Check for $p=7,13$
If $n$ is odd, $8\mid(n^2-1)$
Else $n$ is even, $8\mid n^3$
For $3^2,$ if $3\mid n,3^2\mid n^3$
else $3\nmid n\iff(n,3)=1\implies 9\mid(n^6-1)$ as $\phi(9)=6$
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