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Utomo Theorem ~ The Next Millenium Prize Problem

Utomo Theorem For every prime numbers p which p + 2 is also prime, then \(2^{p+2}\) - 1 always prime.


Can you proves this conjecture?

Note by Budi Utomo
8 months ago

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What you mean is the larger one of every twin prime would produce a Mersenne Prime ? Oh I need to check this out

Aditya Narayan Sharma - 7 months, 1 week ago

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It's actually easier to list the primes \(p\) which satisfy your statement than to find the list that fails it. Just compare any list of twin primes with a list of Mersenne primes, and you'll see that the \(p\) which satisfy your statement are \[p = 3, 5, 11, 17, 29, 59, 1277, 4421, 110501, 132047, \ldots\] and at this point, I got a little tired of looking.

Point being, in the first 8000 or so twin primes, only these 10 satisfy your statement, so there's no way to patch it by removing just a few errant counterexamples.

Brian Moehring - 8 months ago

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you sure that 2^43 - 1 isn't prime number :(

Budi Utomo - 8 months ago

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but, in f(p) = 2^p - 1 hasn't always gives a prime marsenne.

Budi Utomo - 8 months ago

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maybe, except just for p = 41.

Budi Utomo - 8 months ago

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From this theorem, we knew that a large primes is infinite.

Budi Utomo - 8 months ago

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For a counter example, take p=41, 2^43-1 is not prime. See this

Brilliant Member - 8 months ago

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Actually, the numbers of the form 2^n-1 that are prime are known as Mersenne primes. Read this

Brilliant Member - 8 months ago

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