I am currently stuck on a number theory problem. The question is to determine the GCD of every number in the form $p^6-7p^2+6$, where $p$ is a prime number and greater than or equal to $11$.
My approach: After a substitution and factoring I came to factor it as $(p^2-2)(p^2+3)(p-1)(p+1)$. By examining the first $10$ cases I come to the conclusion that the GCD is $672$, as the factors $2^5$, $3$ and $7$ are always present in all the cases. Just I do not know how to prove it.
Yes, $p^6-7p^2+6$ is divisible by $2^5,3,7$ for any prime $p\geq 11$: for $p:=2k+1$, $$\begin{align} p^6-7p^2+6&=(p^2-2)(p^2+3)(p^2-1)\\ &=(p^2-2)(4k^2+4k+4)(4k^2+4k)\\ &=2^4(p^2-2)(k^2+k+1)\underbrace{k(k+1)}_{\text{even}} \equiv 0 \pmod{2^5}, \end{align}$$ and, by Fermat's theorem, $$\begin{align} &p^6-7p^2+6\equiv p^6-p^2=p^2(p^2-1)(p^2+1)\equiv 0 \pmod{3},\\ &p^6-7p^2+6\equiv p^6-1 \equiv 0 \pmod{7}. \end{align}$$ Therefore $2^5\cdot 3\cdot 7=672$ divides $N$, the desired gcd.
Since $11$ and $13$ are primes $\geq 11$, we have that $$\gcd(11^6−7\cdot 11^2+6,13^6−7\cdot 13^2+6)=672$$ is divisible by $N$ ($N$ divides both numbers $11^6−7\cdot 11^2+6$, $13^6−7\cdot 13^2+6$).
Hence we may conclude that $N=672$.