Are there infinitely many primes $p$ of form $$2^k+a^2=p^2<2^{k+2}$$ where $a\in\Bbb N$?

Which primes are known to be of such form?

An example is $16+3^2=5^2$. This is the only one I could find.

  • $\begingroup$ $8+1^2=3^2$. If $0\in\Bbb N$, $4+0^2=2^2$. $\endgroup$ – user236182 Sep 6 '15 at 19:43

If $a=0$ we have the solution $2^2+0^2=2^2\lt2^4$. If $a\gt0$, then

$$\begin{align} 2^k+a^2=p^2&\implies2^k=(p-a)(p+a)\\ &\implies p-a=2^m, p+a=2^n\quad\text{with } m\lt n\text{ and }m+n=k\\ &\implies2p=2^m(1+2^{n-m})\\ &\implies m=1\quad\text{and}\quad p=1+2^{k-2} \end{align}$$

Note that we have $k\gt2$ at this point (since $n\gt m=1$ implies $k=m+n\gt1+1=2$). We now have

$$\begin{align} p^2\lt2^{k+2}&\implies1+2^{k-1}+2^{2k-4}\lt2^{k+2}\\ &\implies2k-4\lt k+2\\ &\implies k\lt6 \end{align}$$

So it remains to check that only $k=3$ and $k=4$ give primes, namely $p=3$ and $p=5$, respectively.


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