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Number Theory Level 5

Consider the sequence a 1 , a 2 , . . . . . . a_{1}, a_{2}, ...... defined by a n = 2 n + 3 n + 6 n − 1 a_{n} = 2^{n} + 3^{n} + 6^{n} - 1 , ( n = 1 , 2 , . . . . ) (n = 1, 2, .... )

Determine all positive integers that are relatively prime to every term of the sequence.

Enter your answer as the total number of integers that you get.


The answer is 1.

Mihir Chakravarti by mihir Chakravarti , 22, India

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1 solution

Mihir Chakravarti
Jan 12, 2015

If p > 3 p>3 , then 2 p − 2 + 3 p − 2 + 6 p − 2 ≡ 1 2^{p-2} + 3^{p-2} + 6^{p-2} \equiv 1 ( m o d p ) (mod p) . To see this multiply both the sides with 6 6 to get: 3 × 2 p − 1 + 2 × 3 p − 1 + 6 p − 1 ≡ 6 3\times2^{p-1} + 2\times3^{p-1} + 6^{p-1} \equiv 6 ( m o d p ) (mod p) , Which is a consequence of Fermat's little theorem. Therefore p p divides a p a_{p} . Also, 2 2 divides a 1 a_{1} and 3 3 divides a 2 a_{2} . So, there is no number other than 1 1 that is relatively prime to all the terms in the sequence.

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