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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.

Bill Dubuque
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Enkt Enktson
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    You don't need to fully factor it. $,f(x) = x^3-7x+6,$ has roots $,x=1,-3,$ so it has factor $,(x-1)(x+3).,$ Putting $x=p^2$ we deduce that $,f(p^2),$ has factor $,(p^2-1)(p^2+3),,$ hence $,8\cdot 4\mid f(p^2),$ by $,8\mid p^2-1,\ 4\mid p^2+3,,$ since ${\rm odd}^2\equiv 1,$ both mod $4$ & $8.,$ See this comment for another way. Finally, the factors $3$ and $7$ follow immediately from Fermat's little Theorem. – Bill Dubuque Aug 26 '23 at 20:34
  • $$\begin{align} &\bmod \color{#0a0}7!:,\ p\not\equiv 0,\Rightarrow, \color{#c00}{p^6\equiv 1}\Rightarrow \ , \color{#c00}{p^6}\ -\ \color{#0a0}7p^2+6\equiv \color{#c00}1-\color{#0a0}0(p^2)+6\equiv 0\[.4em] &\bmod 3!:,\ p\not\equiv 0,\Rightarrow, \color{#c00}{p^2\equiv 1}\Rightarrow (\color{#c00}{p^2})^3!-7\color{#c00}{p^2}!+6\equiv \color{#c00}1^3!-7(\color{#c00}1)+6\equiv 0\ \end{align}\qquad\qquad$$ – Bill Dubuque Aug 27 '23 at 00:49
  • I removed the prime-numbers tag since that is a red herring. The proof works for any set of odd integers not divisible by $7,,$ and includes $11$ & $13$ (or any other index subset yielding $672$ as gcd). – Bill Dubuque Aug 27 '23 at 00:50

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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 $N=672$.

Robert Z
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    You should say why it "suffices to verify that...." – Bill Dubuque Aug 26 '23 at 17:14
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    Your edit did not completely fix the above, since you don't justify why the gcd $M \mid \gcd(f_{11},f_{13}).,$ You are implicitly using a property of gcds that should be made explicit, i.e. you are using w/o justification that $,\gcd(f_{11},f_{13},f_{17},f_{19}\ldots)\mid \gcd(f_{11},f_{13})\ \ $ – Bill Dubuque Aug 26 '23 at 17:27
  • Still incomplete. One straightforward way to deduce that key divisibility is as follows $$m := \gcd(f_{11},f_{13},f_{17},\ldots)\mid f_{11},f_{13},\Longrightarrow, m\mid \gcd(f_{11},f_{13})\qquad\qquad $$ by the gcd universal property $\ \ $ – Bill Dubuque Aug 26 '23 at 17:45
  • @BillDubuque Thanks for your precious comments. OP will appreciate them too – Robert Z Aug 26 '23 at 17:49
  • A bit easier is as follows (this also avoids pulling out of a hat (like magic) the stated factorization of $p^6-7p^2+6).,$ Recall that $\bmod 8!:\ {\rm odd}^2\equiv 1,$ so $,p^2 = 8j+1,,$ so

    $$\begin{align} \bmod 32!:,\ p^2((p^2)^2!-7) &,=,(8j+1)((8j+1)^2!-7)\[.2em] &,\equiv, (8j+1)(16j-6),\ \ \color{lightgrey}{{\rm by},\ \ 8^2\equiv 0}\[.2em] &,\equiv, -6,\ \ \color{lightgrey}{{\rm by},\ \ 8(16)\equiv 0\equiv 1(16)+8(-6)} \end{align}\qquad$$

     – Bill Dubuque Aug 26 '23 at 19:57