If $n\in\mathbb N$ is such that there is only one prime $p$ such that $p^2+2n\in\mathbb P$ and if $m\in\mathbb N$ is such that there is only one prime $q$ such that $q^2+2m\in\mathbb P$, then $3|n-m$. Tested computationally and I would like to see a proof or a counter-example.
Alternative formulation of the test result below: If $n\in\mathbb N$ is such that there is only one prime $p$ such that $p^2+2n\in\mathbb P$, then $n\equiv 1\pmod 3$.
0 value n
: p^2+2n_isprime \ p -- flag
locals| p |
p prime
p p * n 2* + prime
and ;
1 to n { 1 10000 | p^2+2n_isprime } cardinality . 1 ok
2 to n { 1 10000 | p^2+2n_isprime } cardinality . 183 ok
3 to n { 1 10000 | p^2+2n_isprime } cardinality . 125 ok
4 to n { 1 10000 | p^2+2n_isprime } cardinality . 1 ok
5 to n { 1 10000 | p^2+2n_isprime } cardinality . 195 ok
6 to n { 1 10000 | p^2+2n_isprime } cardinality . 221 ok
7 to n { 1 10000 | p^2+2n_isprime } cardinality . 1 ok
8 to n { 1 10000 | p^2+2n_isprime } cardinality . 160 ok
9 to n { 1 10000 | p^2+2n_isprime } cardinality . 329 ok
10 to n { 1 10000 | p^2+2n_isprime } cardinality . 1 ok
11 to n { 1 10000 | p^2+2n_isprime } cardinality . 286 ok
12 to n { 1 10000 | p^2+2n_isprime } cardinality . 129 ok
13 to n { 1 10000 | p^2+2n_isprime } cardinality . 0 ok
14 to n { 1 10000 | p^2+2n_isprime } cardinality . 340 ok
15 to n { 1 10000 | p^2+2n_isprime } cardinality . 232 ok
16 to n { 1 10000 | p^2+2n_isprime } cardinality . 1 ok
17 to n { 1 10000 | p^2+2n_isprime } cardinality . 123 ok
18 to n { 1 10000 | p^2+2n_isprime } cardinality . 182 ok
19 to n { 1 10000 | p^2+2n_isprime } cardinality . 1 ok
20 to n { 1 10000 | p^2+2n_isprime } cardinality . 174 ok
21 to n { 1 10000 | p^2+2n_isprime } cardinality . 295 ok
22 to n { 1 10000 | p^2+2n_isprime } cardinality . 1 ok
23 to n { 1 10000 | p^2+2n_isprime } cardinality . 161 ok
24 to n { 1 10000 | p^2+2n_isprime } cardinality . 249 ok
25 to n { 1 10000 | p^2+2n_isprime } cardinality . 1 ok
26 to n { 1 10000 | p^2+2n_isprime } cardinality . 198 ok
27 to n { 1 10000 | p^2+2n_isprime } cardinality . 129 ok
28 to n { 1 10000 | p^2+2n_isprime } cardinality . 0 ok
29 to n { 1 10000 | p^2+2n_isprime } cardinality . 453 ok
30 to n { 1 10000 | p^2+2n_isprime } cardinality . 362 ok
31 to n { 1 10000 | p^2+2n_isprime } cardinality . 1 ok
I'm using BigZ on https://forthmath.blogspot.se
Anyone is welcome to criticize the test or reformulate the question.
Due to the comment of Ivan Neretin it seems like the following formulation is equivalent:
If $p$ is a unique prime such that $p^2+2n\in\mathbb P$, then $p=3$. And then I belive that the proof essentially is in this proof:
Conditions on $n\in\mathbb N^+$ for $p\in\mathbb P\implies p^2+2n\notin\mathbb P$
