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1. Any Mersenne number $M_p,~p ≡ 3 \pmod 4$ where $p$ is a prime, can be represented as: $$(8px+2p+1)(8py+1) = M_p,~0 ≤ x ≤ \frac{M_p - 1 - 2p}{8p},~ 0 ≤ y ≤ \frac{\frac{M_p}{2p + 1} - 1}{8p}$$

2. Any Mersenne number $M_p,~p ≡ 1 \pmod 4$ where $p$ is a prime, can be represented as: $$(8px+6p+1)(8py+1) = M_p,~0 ≤ x ≤ \frac{M_p - 1 - 6p}{8p},~ 0 ≤ y ≤ \frac{\frac{M_p}{6p + 1} - 1}{8p}$$

Could anyone prove or reject it?

terran
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Nikolay Gladkov
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    If you can prove what you say in your comment, I suggest including it in your question. – coffeemath Jan 28 '24 at 10:40
  • There is more information about it https://docs.google.com/document/d/1LDAJ-mN0rZlJvcnw__WPpvFQqG4MsS4hkemf7nCagac/edit?usp=sharing – Nikolay Gladkov Jan 28 '24 at 10:43
  • Note that in your "factorizations" if one needs to take $y=0$ it makes the second factor $1$ so it is not really a factorization. – coffeemath Jan 28 '24 at 10:44
  • $y$ isn't always should be 0. If y equals 0 we get 1 and it's divisor for every number. $M_{47}$ has divisors like this: $8476 +247 + 1 , 8 47 * 12 + 1 , 8* 47 * 35,278 + 1 $ . We get $x = 6$, $y = 12$, $y = 35,278$ . For prime Mersenne numbers $x = \frac{M_p-1 -2p}{8p},~ y=0$ if $p =~3~(mod~4)$ and $x = \frac{M_p-1 -6p}{8p},~ y = 0$ if $p =~1~(mod~4)$. – Nikolay Gladkov Jan 29 '24 at 04:56
  • In another words the number itself and one. – Nikolay Gladkov Jan 29 '24 at 05:07

1 Answers1

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Yes. Let's work through 1. first. Notice that $p \mid 2^{p-1} -1$ by Fermat's little theorem, so we can write $$2^{p-1} -1 = pm$$ for some $m$. Looking at that equation modulo 4, we see that $-1 \equiv -m \pmod 4$, i.e. $m=4k+1$ for some $k\geq0$.

Now pick $x=k$, $y=0$. We immediately get $$(8px+2p+1)(8py+1) = 2p(4x+1)+1 = 2pm+1 = 2^p -2 +1 = 2^p -1 = M_p.$$ It's pretty straightforward to check that the inequalities hold for $x$ and $y$.

For the second claim, you can construct a similar factorization with $y=0$, I'll leave the proof to you.

marttij
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  • Using Fermat's little theorem seems to be right decision. Why does it nowhere say about it? Everywhere says that you have to looking for divisors in form $2pk + 1$, but it's much harder than looking for divisors in form $8py + 1$ or $8px + 2p + 1$ for $p ≡~3(mod~4)$ and looking for divisors in form $8py + 1$ or $8px + 6p + 1$ for $p ≡~1(mod~4)$. Thank you very much for answer. – Nikolay Gladkov Jan 29 '24 at 04:41