Are there infinitely many $m\in\mathbb N$ such that $35\times2^m+1$ is $\textbf{not}$ a prime number?
Thanks in advance.
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2$\begingroup$ If $m$ is even the number is divisible by $3$. $\endgroup$– Dag Oskar MadsenCommented Oct 13, 2020 at 9:10
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1$\begingroup$ In general, it is easy to see that for any $a,b,c\in\mathbb Z$ such that $a\ne0$ and $b\ne0,\pm1$, there is an infinite arithmetic progression of $m$’s such that $ab^m+c$ is composite. $\endgroup$– Emil JeřábekCommented Oct 13, 2020 at 9:51
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2$\begingroup$ Any reason for $35$ in particular? $\endgroup$– WojowuCommented Oct 13, 2020 at 10:17
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2 Answers
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Yes.
For natural $k$ let $m=4+10k$. Then $35 \times 2^m+1$ is divisible by $11$, since $35 \cdot 2^{4+10k} \equiv -1 \pmod {11}$ and $2^m$ is periodic modulo all primes.
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For any positive integer $m$, there are infinitely many $n$ such that $35 \times 2^n+1$ is divisible by $35 \times 2^m+1$. Indeed, it suffices to take $n$ such that $2^{n-m} \equiv 1 \mod 35 \times 2^m+1$.
answered Oct 13, 2020 at 16:55