10
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How can I prove in general that, for all $n\geq 2$:

$$ \gcd(p_{n-1}, \ n^5 - n^3 + n^2 - n + 1) = 1 $$

Seems to always be true:

from sympy import *
from sympy.ntheory.generate import prime


def f(n):
    return n**5 - n**3 + n**2 - n +  1

for n in range(3, 100000000):
    p = prime(n-1)
    d = gcd(f(n), p)
    if d != 1:
        print (n)

Assume true.

$$ f(X) = X^5 - X^3 + X^2 - X + 1 $$

is irreducible and modulo each prime, i.e. $f(n+1) \neq 0 \pmod {p_n}$ so that $\prod_{j=2}^n\overline{f(j+1)} \in $ the units of $\Bbb{Z}/2 \times \Bbb{Z}/3 \times \Bbb{Z}/5 \times \dots $

So far it seems like the polynomials $g(X) = f(X \pm 1)$ are such that $g(n) \neq 0 \pmod {p_n}$

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  • 6
    $\begingroup$ I see this as a coincidence (what's the chance of the polynomial evaluated at n divisible by the n-th prime?). $\endgroup$ – Kenny Lau Sep 27 '17 at 9:43
  • $\begingroup$ @KennyLau try other polynomials with the code. They don't show the same result. $\endgroup$ – BananaCats Category Theory App Sep 27 '17 at 9:45
  • 1
    $\begingroup$ For example?$ $ $\endgroup$ – Kenny Lau Sep 27 '17 at 9:56
  • 4
    $\begingroup$ Why calculate the gcd? $p_{n-1}$ is a prime, it's a divisor of $f(n)$ or not. $\endgroup$ – Professor Vector Sep 27 '17 at 10:05
  • 2
    $\begingroup$ i did not solve it in the comment nor mentioned any thing useful to a proof, in general most of the questions in NT that include both $p_n,n$ are hard to solve and the majority of these question are open problems ,because $p_n$ behave an a very irregular way relevant to $n$. $\endgroup$ – Ahmad Sep 27 '17 at 10:41

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