0
$\begingroup$

Conjecture:

Any odd natural number $n\notin \{1,27\}$ is of form $n=a+b,\,a,b\in\mathbb N^+$, where $a^2+b^2$ is a twin prime.

This is a stronger variant of the conjecture Any odd number is of form $a+b$ where $a^2+b^2$ is prime and I would like help with finding counterexamples or test limits.

So far I've tested for $n<1,000,000$.


I guess there is a good chance for a counterexample and I offer an award of 500 points for the first posted counterexample.

Of course I would like valid heuristic arguments for the conjecture, even if I don't believe they exists.


Correction:
As it seems, there exist an unlimited number of counterexamples in the first statement, but not in the second statement:

  1. Any odd natural number is of form $n=a+b,\,a,b\in\mathbb N^+$, where both $m=a^2+b^2$ and $m-2$ are primes.

  2. Any odd natural number is of form $n=a+b,\,a,b\in\mathbb N^+$, where both $m=a^2+b^2$ and $m+2$ are primes.

$\endgroup$
  • 1
    $\begingroup$ If there are only finitely many twin primes, there'll be only finitely many pairs $(a,b)$, so your conjecture implies the twin prime conjecture. $\endgroup$ – Barry Cipra Sep 9 '17 at 13:27
  • $\begingroup$ is n a natural number ? or an integer ? $\endgroup$ – user451844 Sep 9 '17 at 13:27
  • $\begingroup$ Are $a$ and $b$ also natural numbers, or can you have, for example $n=7=10-3$ where $10^2+(-3)^2=109$? $\endgroup$ – Barry Cipra Sep 9 '17 at 13:33
  • 1
    $\begingroup$ @michaelmross, $5=2^2+1^2$ gives $n=2+1=3$; $13=3^2+2^2$ gives $n=3+2=5$, as does $17=4^2+1^2$; $29=5^2+2^2$ gives $n=5+2=7$; etc. In any twin prime pair, one of the primes is congruent to $1$ mod $4$, hence the sum of two squares. $\endgroup$ – Barry Cipra Sep 9 '17 at 13:46
  • 2
    $\begingroup$ It has been shown that the number of twin primes $< N$ is bounded above by $\frac{CN}{(\log N)^2}$, so the number of ways to write $a+b$ where $a^2+b^2$ is a twin prime is asymptotically $\leq \frac{N^2}{(\log N)^2}$. Heuristically, this means that the best known bound on twin prime density doesn't come close to hinting at the existence of a counterexample. $\endgroup$ – Carl Schildkraut Sep 9 '17 at 16:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.