7
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
2
$\begingroup$

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 $(A_1q_1^n+A_2q_2^n)$.

$\endgroup$
  • $\begingroup$ $a \mid b$ $a \mid b$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.