How can I see if $2^{333}-1 $ is a prime number?
Does this have to do with Mersenne prime numbers ($2^n-1$) ??
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closed as too localized by Thomas, mikeazo♦ Jan 31 at 21:42
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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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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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