Let $q_1,\cdots,q_m$ be distincts elements of $\mathbb N\setminus\{0,1\}$. Does there exist non zero integers $A_1,\cdots,A_m$ such that infinitely many primes divide no elements of the sequence $\left(A_1q^n_1+\cdots+A_mq_m^n\right)_{n\in\mathbb N}$?

Any aid or answer would be welcome. Thanks in advance.

  • $\begingroup$ I believe the answer is no if the q_i are sufficiently distinct. I will see I can find a reference. Gerhard "Checking On My Advisor's Work" Paseman, 2019.12.08. $\endgroup$ – Gerhard Paseman Dec 9 '19 at 2:59
  • $\begingroup$ The answer to the literal question is "yes" with $A_1 = 1$ and $A_2 = \dotsc = A_m = 0$. So please disallow $A_i = 0$ or I will post this answer... $\endgroup$ – WhatsUp Dec 9 '19 at 4:36
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    $\begingroup$ It is believed that up to simple obstructions, if for one $p$ the sequence is coprime with $p$ then the same happens for infinitely many $p$. $\endgroup$ – reuns Dec 9 '19 at 17:33

You can prove this relatively easily for $m=2$.

Chose $A_1, A_2$ so that there exist infinitely many primes with $q_1q_2$ a quadratic residue and $-A_1A_2$ a non-residue mod $p$. This can be done using quadratic reciprocity and gives the primes lying in an arithmetic progression. (For example if $p\equiv -1 \bmod 24$ then 2 is a residue and -3 is a non residue mod $p$.)

Then $p\mid(A_1q_1^n+A_2q_2^n)$ means that $A_1q_1^n+A_2q_2^n\equiv0 \bmod p$ or $A_1^2q_1^{2n}+A_1A_2{(q_1q_2)}^n\equiv0 \bmod p$ which cannot be true if $-A_1A_2$ is a non residue and $q_1q_2$ is a residue.

Hence all such $p$ cannot divide $(A_1q_1^n+A_2q_2^n)$.

  • $\begingroup$ $a \mid b$ $a \mid b$ and $a \equiv b \bmod n$ $a \equiv b \bmod n$ give better spacing than $a|b$ $a|b$ and $a \equiv b$ mod $n$ $a \equiv b$ mod $n$. I have edited accordingly. $\endgroup$ – LSpice Dec 17 '19 at 15:17

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