When trying to solve a bigger problem, I came across the problem of characterising all primes $p,q$ such that $p^2+1=2q$. That is, are there necessary and sufficient conditions for primes $p,q$ to satisfy the equation? Even better, is there a paramatrisation of solutions (though, this is probably unlikely)? I know the solutions $(p,q)=(3,5),(5,13),(11,61),(19,181),(29,421)$ but I don't know that these are the only ones (indeed, it seems likely there are many more).

Some incomplete thoughts: I immediately made the factorisation $(p+1)(p-1)=2(q-1)$, and because $p$ is obviously odd, let $p=2p_1+1$. This leads us to the equation $2p_1(p_1+1)=q-1$. If we substituted $q=2q_1+1$ we have $p_1(p_1+1)=q_1$, so $q$ is twice of the product of two consecutive integers, plus one. It follows that $q$ is $1$ mod $4$, and $p^2$ is $1$ mod $8$. The same conclusion can be obtained simply by noting that $-1$ must be a quadratic residue mod $2q$, hence $1$ is a $4$th power residue mod $2q$. By Lagrange's Theorem we have $4\mid\phi(2q)=q-1$, so $q \equiv1$ mod $4$. Any further thoughts are appreciated!

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    $\begingroup$ oeis.org/A048161 $\endgroup$ – mathlove Dec 29 '18 at 5:51
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    $\begingroup$ You should definitely look at the link in the first comment. Whether there are infinitely many such $p$ is unknown... BTW if $p>5$ then $p\equiv \pm 1 \mod 10,$ for if $5<p\equiv \pm 3 \mod 10$ then $ ( p^2+1)/2$ is greater than $5$ and divisible by $5.$ $\endgroup$ – DanielWainfleet Dec 29 '18 at 6:28
  • $\begingroup$ @mathlove Wow, thanks. If you make that an answer I would happily accept it :) $\endgroup$ – YiFan Dec 29 '18 at 8:50
  • $\begingroup$ @YiFan Do you want to find all the solutions or only decide whether infinite many exist ? Besides some basic restrictions, there won't be useful methods. You will just have to check the number $\frac{p^2+1}{2}$ in general. If you are interested in an evidence that infinite many solutions probably exist, I can undelete my answer. $\endgroup$ – Peter Dec 30 '18 at 10:44
  • $\begingroup$ @Peter I wanted to see if there was a nice necessary and sufficient criterion that could be used to describe all solutions to the equation, but I now know that's probably not going to happen. I'm certainly interested in anything related to the problem though, and your answer is welcome :) $\endgroup$ – YiFan Dec 30 '18 at 11:21

We have infinitely many primes of the form $$\frac{p^2+1}{2}$$ where $p$ is itself prime, if we have infinite many positive integers $k$ such that $$2k+1$$ and $$2k^2+2k+1$$ are simultaneously prime. (in this case, just set $p=2k+1$). The Bunyakovsky conjecture implies that this is the case, so very likely infinitely many examples exist. But I am convinced that the problem is open.


I propose the following "kind of parametrization". Remark first, as you did, that necessarily $q\equiv 1$ mod $4$ (this can be seen directly from the characterization of integers which are sums of squares, see e.g. Samuel's ANT, V-6). Then work in the Gauss ring $\mathbf Z[i]$, which is a PID with units $u=\pm 1,\pm i$. For convenience, write $z'≈z$ for $z'=uz$. It is classically known that $2$ is totally ramified and $q$ splits completely in the Gauss ring, more precisely $2=(1+i)^2≈(1+i)(1-i)$ and $q≈(x+iy)(x-iy)$, where $x\pm iy$ are prime elements of $\mathbf Z[i]$. Because of unique factorization (up to units), the given equation will be equivalent to $p+i≈(1+i)(x+iy)$, hence our "pseudo-parametrization" will be as follows: 1) Take a prime $q\equiv 1$ mod $4$ and decompose it in $\mathbf Z[i]$; 2) Solve $p+i≈(1+i)(x+iy)$ in $\mathbf Z^2$. Because of symmetry, let us solve only the equation $p+i=-(1+i)(x+iy)$. By identification, $p=y-x$ and $-1=x+y$, so $p=2y+1$. It remains however to check that $2y+1$ is a prime in $\mathbf Z$, which is why our method cannot be considered as a genuine parametrization. Although we have just shown that the OP question is essentially equivalent to the Bunyakovsky conjecture evoked by @Peter, I don't know if this gives any new hint.


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