Primes of the form $(2m+1)^2-2^{2s+1}$

The question is the following :

• Question:

Does there exist infinitely many primes of the form $(2m+1)^2-2^{2s+1}$ with $m,s\geq 1$ ?

• Why this could be true:

Bunyakowsky conjecture would imply that the answer is yes for any given $s$.

[ Indeed, the polynomial $4X^2+4X-2^{2s+1}+1$ is irreducible ($\Delta = 2^{2s+5}$), and no integer $d\geq 1$ can divide all values $P(n)$ (since $P(0)$ is odd and $P(1)-P(0)=8$) ]

Unfortunately, as the Wikipedia page indicates, Bunyakowsky conjecture is not known to hold for a single polynomial of degree $\geq 2$.

My hope is that allowing $s$ to assume infinitely many values can help ...

• Why this could be hard:

As was pointed out by Mike Benett in this MO question, the density of primes of this form among primes that are congruent to $1$ mod $8$ is asymptotically $0$. (As an example, there are $20453$ primes $p<2^{20}$ that are congruent to $1$ mod. $8$, and only $1334$ among them that have the required form, so that the density up to that bound is only $0.065$).

• The conjecture that there are infinitely many primes of form $(2m)^2+1$ is still open (and was listed as one of Landau's four problems more than 100 years ago), so maybe this is another indication your question could be hard? – Jeppe Stig Nielsen Oct 14 '15 at 19:19
• Do you also count negative prime values of your expression $(2m+1)^2 - 2^{2s+1}$? – Stefan Kohl Oct 14 '15 at 20:12
• @JeppeStigNielsen : Right ... as far as I know, Euler had already asked this very question. – few_reps Oct 14 '15 at 21:53
• @StefanKohl : definitely not ! The point is that these primes furnish interesting real quadratic extensions. – few_reps Oct 14 '15 at 22:02
• I think proving this would be a fantastic achievement. Compare it with the $x^2+y^4$ result of Friedlander-Iwaniec, and note that there are much more fourth powers than powers of two. – GH from MO Oct 14 '15 at 22:22