mersenneforum.org Finding all divisors kn + 1 of P(n) for various polynomials P
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2014-11-28, 14:51   #1
Drdmitry

Nov 2011

22×3×23 Posts
Finding all divisors kn + 1 of P(n) for various polynomials P

Let $P(x)\in Z[x]$ be a monic polynomial with such that $x^nP(-x^{-1}) = P(x)$. I was interested in looking at the divisors of the values P(n) of the form kn+1. It appears that, given one divisor
$n_0n_1+1 | P(n_1),$
there is an infinite series of the divisors of this form which is given by the equations
$P(n_k) = (n_{k-1}n_k+1)(n_{k+1}n_k+1).$
For the polynomial $P(n) = n^4 +$1 one can classify all such series. They are "generated" by the pairs $(n_0, n_1) = (0, n)$ for an arbitrary $n\in N$. This in turn implies (with some efforts to be made) that numbers $b^{4m}+1$ do not have (non-algebraic) divisors of the form $kb^m + 1$.

Concerning the polynomial $P(n) = n^8 + 1$ we also have series of divisors of P(n) generated by the pairs $(n_0, n_1) = (0,n)$. Additionally the pairs $(n_0, n_1) = (n^3, n^5)$ generate infinite series of divisors of P(n). However a basic search among small numbers shows that there are still "exceptional" pairs $(n_0,n_1)$ which generate the divisors $n_0n_1+1 | n_1^8 + 1$, the smallest of them is (3,11). With some efforts one can check that all (non-algebraic) divisors $kb^m+1$ of numbers $b^{8m}+1$ must come from an "exceptional" pair $(k,b^m)$.

I do not know how to classify the "exceptional" pairs $(n_0, n_1)$. I conducted a search for all pairs with $\max\{n_0,n_1\} \le 10^6$ and additionally with $n_0+n_1\le 10^7$. In total there are 201 different infinite series found. Also it seems that they are more less equidistributed on a $\log n_0 \times \log n_1$ coordinate plane.

It would be very interesting to find a way to classify all of the exceptional pairs $(n_0,n_1)$. In particular it may give us all divisors $kb^m + 1$ of numbers $b^{8m}+1$.
Attached Files
 spec_fact.pdf (199.9 KB, 192 views) x^8+1_chains.txt (10.3 KB, 256 views)

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