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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 $p\in\mathbb P\implies p^2+2n\notin\mathbb P$

$\endgroup$
  • $\begingroup$ I don't really understand what you're asking with $n = a$ and $n=b$ $\endgroup$ – mdave16 Mar 4 '17 at 21:59
  • $\begingroup$ but what is $n$? since you've defined it twice $\endgroup$ – mdave16 Mar 4 '17 at 22:06
  • $\begingroup$ nvm, the answer below makes it clearer, i figured out the question $\endgroup$ – mdave16 Mar 4 '17 at 22:07
  • 2
    $\begingroup$ Quite obviously, if $n\equiv 1\pmod 3$, then $p^2+2n$ is divisible by 3 for all prime $p$ except 3, and hence not prime. As for the proof in the other direction, it may pretty well be not known yet. $\endgroup$ – Ivan Neretin Mar 6 '17 at 21:34

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