divisibility of $n^{15} - n^3$ by $32760$

Question

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 $n \ge 0$.

Answer 1

$32760=2^3\cdot3^2\cdot5\cdot7\cdot13$

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 $\phi(9)=6$

Answer 2

A much less elegenat approach: If $f$ is a polynomial of degree $d$ and $f(0),\ldots, f(d)$ (or any $d+1$ consecutive values) are multiples of an integer $m$, then $f(n)$ is dividible by $m$ for all integers $n$. (This statement can be shown by induction on $d$ using repeated differences). So without knowing a factorization of $32760$, you could "simply" test the claim for $n=0,1,\ldots, 15$ (or $-7,-6,\ldots,8$).

Answer 3

$$ \begin{align} \mathbb{Z}_{32760}^{\times} &\cong\mathbb{Z}_{8}^{\times}\times\mathbb{Z}_{9}^{\times}\times\mathbb{Z}_{5}^{\times}\times\mathbb{Z}_{7}^{\times}\times\mathbb{Z}_{13}^{\times}\\ &\cong \left(C_2\times C_2\right)\times C_6\times C_4\times C_6\times C_{12} \end{align}$$

So for any $n$ relatively prime to $32760$, $n^{12}\equiv 1$ mod $32760$. So $n^{15}\equiv n^3$.

If $n=ab$ where $b$ is maximally relatively prime to $32760$, then just as above, $b^{15}\equiv b^3$ mod $32760$.

So it remains to consider $a$ where $a=2^{s_2}3^{s_3}5^{s_5}7^{s_7}13^{s_{13}}$. So $a^3=2^{3s_2}3^{3s_3}5^{3s_5}7^{3s_7}13^{3s_{13}}$

There is an isomorphism $$\mathbb{Z}_{32760} \cong\mathbb{Z}_{8}\times\mathbb{Z}_{9}\times\mathbb{Z}_{5}\times\mathbb{Z}_{7}\times\mathbb{Z}_{13}$$ which can be factored as $$\mathbb{Z}_{32760} \cong G_0\times G_1$$ where $G_0$ has the factors corresponding to primes where $s_i>0$, and $G_1$ has the factors corresponding to primes where $s_i=0$. Projecting $a^3$ onto $G_0$ gives $0$, and projecting $a^3$ onto $G_1$ gives a unit of $G_1$. So as with $n$ that are relatively prime to $32760$, in $G_1$ $a^{12}\equiv1$. Either way, $a^{15}\equiv a^3$.