Prime as a sum of Geometric Series

Let x and n be positive integers such that \(1 + x + x^2 + … + x^{n-1} \) is a prime number. Then show that n is a prime number.

Solution:

(For small values of x and n it is easy to show that the given fact is true. We prove for x>>1)

Suppose n is not a prime. Then n = ab (where both a and b are not equal to 1). We may write the given expression in blocks of a terms; there will be b such blocks.

prime = \(1 + x + x^2 + … + x^{n-1} = (1 + x + x^2 + … + x^{a-1} ) + x^a (1 + x + x^2 + … + x^{a-1} ) + x^{2a} (1 + x + x^2 + … + x^{a-1} ) + … + x^{a(b-1)} (1 + x + x^2 + … + x^{a-1} ) \)

=\((1+x+x^2 +…+x^{a-1}) (x^a + x^{2a} +…+ x^{a(b-1)}) \)

But this gives a factorization of a prime number which is not possible (as x>>1 none of the factor equals 1). Hence we find a contradiction implying n is prime.

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