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Conjecture:

For each prime $p$ there are an infinite number of primes $q$ such that $p+q$ is a perfect square.

I have done a lot of tests using Bigz and I believe that it's possible to prove.

I change from possible to prove to not possible to disprove.
Define for all primes $q$, $w_q(n)=|\{p\in\mathbb P|p<n\wedge \exists k\in\mathbb N:p+q=k^2\}|$.
Then I've observed that it seems that for all primes $q$ there exists a natural number $N$ and a $\alpha\in\mathbb R_+$ such that $m>N$ implies that $w_q(10^m)>\alpha2^m$. This would imply the conjecture and is maybe possible to disprove.

Lehs
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2 Answers2

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This would imply that for every prime $p$ , there are infinitely many primes of the form $n^2-p$ which is a well-known (open) problem.
It is a well known conjecture that if $m$ is not a perfect square, then there are infinitely many primes in $n^2-m$ , $n\in \mathbb{N}$ without saying anything if $m$ is prime.

Do not expect much.

Konstantinos Gaitanas
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This is not an answer (or perhaps it is, if little more can be said), but too long for a comment.

As far as I know:

Dirichlet's theorem implies that the sequence $\{an-p\}_n$ has infinitely many primes if $a>0$ and is not a multiple of $p$. Your question changes $an$ for $n^2$. This seems much more difficult. Dirichlet's theorem's proof is based in the fact that $$\sum_{an-p\text { prime}}\frac1{an-p}$$ diverges. But with $n^2$ the series converges, so we can't say (based only on this) if the sum has actually infinitely many terms.

I'm afraid that it would be very hard to prove that, for every prime $p$, the sequence $\{p+q\}_{q\text{ prime}}$ has at least a single square at all.

As an example, see this.

ajotatxe
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