Question

Prove that for every prime $p>7,p^6-1$ is divisible by $504.$

Answer 1

Since $504=2^3{.3}^2.7,$ therefore it is enough to show that $p^6-1$ is a multiple of $2^3,3^{2}and7$.
Step 1: Divisibility by 8:
$p^6-1$ is divisible by $p^2-1$. Since p is a prime greater than 7, therefore $p-1$ and $p+1$ are both even. Out of the consecutive even integers $p-1$ and $p+1$, one must be a multiple of 2 and the other must be a multiple of 4. (In fact, if $p-1=4k+2,$ then $p+1=4k+4;$
if $p-1=4k,thenp+1=4k+2$. In either case $(p-1)(p+1)$ is a multiple of 8).
Step 2: Divisibility by 9:
Since the product of three consecutive integers $p-1,p,p+1$ is divisible by 3,and p is not divisible by 3 (because it is a prime greater than 3), therefore $(p-1)(p+1)$ is divisible by 3.
Now $p^6-1=(p^2-1)(p^4+p^2+1),$
$=(p^2-1)\left\{\right(p^2-1)+3p^2\}$
Since $p^2-1$ is divisible by 3, therefore $(p^2-1{)}^2+3p^2$ is also divisible by 3. Consequently $p^6-1$ is divisible by 9.
Step 3: Divisibility by 7:
Since 7 is prime 7 and p is prime to 7 (being a prime greater than 7), therefore by Fermat's theorem $p^6-1$ is multiple of 7.
Since $p^6-1$ is divisible by 8, 9 and 7 and the numbers 8, 9 and 7 are co-prime, therefore it is divisible by $8\times 9\times 7,i.e.,504$