A prime a 2 b − 1 a^{2b} - 1

Number Theory Level 1

True or False?

There exists a prime number a 2 b − 1 a^{2b} - 1 , for which a a and b b are integers, with a a as an odd number.

True False

Ron Lauterbach by Ron Lauterbach , 18, Germany

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3 solutions

Jayden Johnson
Sep 21, 2017

Notice that the expression above can be written as the difference of squares: (a^b)^2-1^2. This is equivalent to (a^b-1)(a^b+1), and therefore cannot be prime.

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Jayden, it could be a prime if a^b -1 = 1, so we have to exclude the possibility that a^b could = 2. It is the fact that a is odd that makes it impossible. Just a small point

Edwin Gray - 2 years, 3 months ago
Edwin Gray
Mar 2, 2019

a^(2b) - 1 = [a^b]^2 - 1 = (a^b - 1)(a^b + 1. If prime, a^b - 1 = 1, or a^b = 2, which cannot be for a odd.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

a^2b has to be an odd number and if you -1 from that, it will be an even number so no. It's not possible.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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