Determine the greatest common divisor of the numbers $p^6-7p^2+6$ where $p$ runs through the prime numbers $p \ge 11$.
Problem
Source: Bundeswettbewerb Mathematik 2023, Round 2 - Problem 1
Tags: number theory, greatest common divisor, number theory proposed, GCD, prime numbers
iniffur
08.09.2023 17:52
My try:
$f(p)=p^6-7p^2+6$
$f(\pm1)=0\Rightarrow p^6-7p^2+6=(p+1)(p-1)(p^4+p^2-6)$
$p^2=X\rightarrow p^4+p^2-6=X^2+X-6=(X+3)(X-2)$
$\Longrightarrow p^6-7p^2+6=(p-1)(p+1)(p^2+3)(p^2-2)$
$p=4k+1\Longrightarrow (p-1)(p+1)(p^2+3)(p^2-2)=32k(2k+1)(4k^2+2k+1)(16k^2+8k-1)$
$p=4k-1\Longrightarrow (p-1)(p+1)(p^2+3)(p^2-2)=32k(2k-1)(4k^2-2k+1)(16k^2-8k-1)$
Therefore, the greatest common divisor of the numbers $p^6-7p^2+6$ when $p$ runs through the prime numbers $p \ge 11:$ $$32$$.
@below: fixed, Thx
Ianis
08.09.2023 18:46
ddot1 wrote:
$(p-1)(p+1)(p^2+3)(p^2-2)$
? Yes, then observe that $(p-1)(p+1)\equiv 0\pmod 8$ and $p^2+3\equiv 0\pmod 4$ for all primes $p\geq 11$, so the expression is always divisible by $2^5$. Using Fermat's Little Theorem we find that $p^6-7p^2+6$ is divisible by $3$ and $7$. Evaluating at $p=11$ and $p=13$ we conclude that the gcd is $2^5\cdot 3\cdot 7$.
Math4Life7
08.09.2023 18:55
You can just note that you only have to care about $v_2$, $v_3$, $v_5$, and $v_7$ and you can do those cases manually very easily