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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.

Lehs
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    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. – Barry Cipra Sep 09 '17 at 13:27
  • is n a natural number ? or an integer ? –  Sep 09 '17 at 13:27
  • Are $a$ and $b$ also natural numbers, or can you have, for example $n=7=10-3$ where $10^2+(-3)^2=109$? – Barry Cipra Sep 09 '17 at 13:33
  • Thanks! I've corrected the conditions. – Lehs Sep 09 '17 at 13:37
  • Can you provide a short list of examples? I'm not clear what you're saying. The most intuitive thing you can say about the relationship of twin primes and perfect squares is each twin prime pair are the prime factors of $x^2-1$? – michaelmross Sep 09 '17 at 13:39
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    @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. – Barry Cipra Sep 09 '17 at 13:46
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    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. – Carl Schildkraut Sep 09 '17 at 16:59
  • @CarlSchildkraut, what I was thinking of was that it might be difficult to show heuristic arguments that the conjecture is true. – Lehs Sep 09 '17 at 17:05
  • Your questions are not turned into math questions. You should read at least Apostol's book. The random model for the primes is that the random variables $X_n = 1_{n \text{ is prime}}$ are independent with $P[X_n = 1] = \frac{1}{\log n}$. Thus $P[X_n X_{n+2} = 1] =P[X_n \land X_{n+2} = 1] = \frac{1}{\log n} \frac{1}{\log (n+2)} \sim \frac{1}{\log^2 n}$ and you want to look at the mean and variance $\mu_n,\sigma_n^2$ of $Y_n=1- \prod_{a= 2}^{n-2} (1-X_n X_{n+2})$. – reuns Sep 09 '17 at 17:30
  • If $\lim_{n \to \infty} \frac{\sigma_n}{\mu_n} = 0$ then under the random model, the probability that there is no counter-example for $n$ large enough is $1$. @CarlSchildkraut – reuns Sep 09 '17 at 17:30
  • What if you remove the word "twin" from the conjecture and include the case $n=27$ (because $27=26+1$ and $26^2+1^2=677$ is prime)? Is every odd $n>1$ equal to $a+b$ where $a,b$ are positive integers and $a^2+b^2$ is prime? – DanielWainfleet Sep 09 '17 at 18:50
  • @DanielWainfleet: https://math.stackexchange.com/questions/2386860/any-odd-number-is-of-form-ab-where-a2b2-is-prime?rq=1 – Lehs Sep 09 '17 at 19:10

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