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Old 2020-06-20, 07:55   #6
JeppeSN
 
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"Jeppe"
Jan 2016
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If you consider the numbers 2^(4k+2) + 1, there are two kinds of algebraic factorizations for them. One is because 4k+2 = 2(2k+1), so the exponent has an odd divisor, therefore:

2^(4k+2) + 1 = (2^2)^(2k+1) + 1 = (2^2 + 1)[(2^2)^(2k) - (2^2)^(2k-1) + ... - (2^2)^3 + (2^2)^2 - 2^2 + 1]

But this simply says that 2^2 + 1 = 5 divides 2^(4k+2) + 1 which is also quite obvious without the factorization above.

Set n=2k+1. When n is a prime, there are no other nontrivial odd factors of the exponent 4k+2 than n, and then we do not get more factorizations of the above type.

However, Aurifeuille noted another algebraic factorization of 2^(4k+2) + 1, namely

2^(4k+2) + 1 = (2^(2k+1) - 2^(k+1) + 1)(2^(2k+1) + 2^(k+1) + 1)

For each k, one of the two factors here is necessarily divisible by 5. Then the other may be prime.

We can also write Aurifeuille's factorization with j = k+1 instead:

2^(4j-2) + 1 = (2^(2j-1) - 2^j + 1)(2^(2j-1) + 2^j + 1)

or with n (which is odd):

2^(2n) + 1 = (2^n - 2^((n+1)/2) + 1)(2^n + 2^((n+1)/2) + 1)

Let n be this odd number. The relation to Mersenne primes, as explained on Caldwell's page linked by carpetpool and Batalov above, is that the complex number (1+i)^n - 1 where i is a square root of minus one, is a prime in the ring of Gaussian integers exactly if the one of Aurifeuille's factors which is not a multiple of 5, is an ordinary prime.

The classical (https://en.wikipedia.org/wiki/Aurife...zation#History) example is n = 29 (so k=14 and j=15) where Aurifeuille says:

2^58 + 1 = (2^29 - 2^15 + 1)(2^29 + 2^15 + 1)

where the latter factor is prime (so a Gaussian Mersenne norm prime corresponding to the Gaussian prime (1+i)^29 - 1 where i is either of the two square roots of minus one).

/JeppeSN
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