Number theory primes and congruences: $p^{32} -1$ is divisible by $16320$ I had problem in the following problem any help or hint will highly be appreciated.  
For $p>17$, $p^{32} -1$ is divisible by $16320$
 A: Hint: we have $16320=2^6\cdot 3\cdot 5\cdot 17$, and already $p^{32}\equiv 1 \bmod k$ for each $k\in \{2^6,3,5,17\}$. For example, $p^2\equiv 1\bmod 3$, because $\gcd(p,3)=1$.
A: You have $16320=2^6\cdot3\cdot5\cdot17$.
Notice that using Fermat's theorem you have:


*

*$p^{16}\equiv1\pmod{17}$, hence  also $p^{32}=(p^{16})^2\equiv1\pmod{17}$

*$p^4\equiv1\pmod5$, hence also $p^{32} \equiv 1 \pmod5$

*$p^2\equiv1\pmod3$. hence also $p^{32} \equiv 1\pmod3$


Now it only remains to notice that $\phi(64)=32$ and by Euler's theorem $p^{32}\equiv1\pmod{64}$.
And now you can combine the above congruences together.
A: NOTE.  It would be absurd that this property is valid for all prime number greater than $17$; indeed $29^{32}-1=(29^{16}-1)(29^{16}+1)$. Calculating we have
$29^{16}=250246473680347348787 521$ and it is verify that neither $29^{16}-1$ nor $29^{16}+1$ is divisible by $17$ hence not divisible by a multiple of $17$ thus not divisible by 16320.
NOTE (bis).The first note was a mistake induced by both, certain idea about the anarchy of primes and the calculator. It is so clear that the property of divisibility by $17$ is true……. The "hidden" $16 = 17-1$, which could be applied to any prime p with the exponent appropriate  $2(p-1)$..... But this usually happens. See for example my answer to "Can a pre-calculus student prove this?" and compare it with the other ones. Here also the answer is immediate but qualified people did not see.   
Therefore the question must be understand as looking for a prime satisfying the enunciated property. I find below the prime $p=16319$
The number $p=16320-1=16319$ is prime and $p^{32}=16320M+1$ for some $M$. Hence you have $p^{32}-1$ is a multiple of $16320$.
Note anyway that for all even k you have the same result.
