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How can I see if $2^{333}-1 $ is a prime number?

Does this have to do with Mersenne prime numbers ($2^n-1$) ??

Thank you!

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closed as too localized by Thomas, mikeazo Jan 31 '13 at 21:42

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Ask Wolfram|Alpha! –  Thomas Jan 31 '13 at 21:13
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2 Answers 2

up vote 5 down vote accepted

No, $2^{333} - 1$ cannot be a prime. It is easy to see this via the exponent, $333$, which has factorization $3\cdot3\cdot37$. From this we know, by a well-known identity, that

$$ 2^{333} - 1 = (2^{37} - 1)\cdot (1 + 2^{37} + 2^{37\cdot2} + 2^{37\cdot3} + 2^{37\cdot4} + 2^{37\cdot5} + 2^{37\cdot6} + 2^{37\cdot7} + 2^{37\cdot8}). $$

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It's also divisible by $2^3 - 1 = 7$ by the same argument. i.e. It's not even close to prime. –  Joe Z. Feb 1 '13 at 20:38
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No it's not prime. Its prime factorization (in ascending order) is

$7×73×223×1999×10657×169831×321679×1238761×26295457×36085879×199381087×319020217×616318177×698962539799×4096460559560875111$

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Source:mathematica (and wolframalpha can do it too) –  AFS Jan 31 '13 at 21:20
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