Prove that if $\frac{3^{n}-1}{2}$ is a prime number then so is n

Question

How can I prove if $\frac{3^{n}-1}{2}$ is a prime number then so is n. I tried concluding that $3^{n}-1$ needs to be in the form of 2×p (p is a prime number) so $\sum_{i=0}^{n-1} 3^n$ is a prime number, but I can't think of a way to prove that either.

Welcome to Math SE. FYI, using an Approach0 search, there's the similar Prime numbers of the form $(3^n-1)/2$, where the first part of its currently accepted answer uses a similar argument to that in this answer here. The search also comes up with the related Is there a Lucas-Lehmer equivalent test for primes of the form ${3^p-1 \over 2}$?. — John Omielan, May 03 '23 at 20:32

score -1 · Accepted Answer · edited Jun 07 '23 at 11:35

-1

Hint: if $n = ab$ is composite, then you can write $$3^n - 1 = (3^a - 1)(3^{(b - 1)a} + 3^{(b - 2)a} + \dots + 3^a + 1).$$

edited Jun 07 '23 at 11:35

Dietrich Burde

130,978

answered May 03 '23 at 20:23

CJ Dowd

1,664

1

Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque May 03 '23 at 20:58

Prove that if $\frac{3^{n}-1}{2}$ is a prime number then so is n

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