# Find all primes of the form $3^n -1$

Regarding the question, the prime must be a natural number, so n>0. I used Mathematica to test for primes for n ranging from $1$ to $10^5$. All of them were not prime aside from $n=1$, which gives you $2$. I believe contradiction would be used for proving that there are no other primes of this form. I am just not sure where to go from there.

• Well, $3^n-1$ is even, right? – carmichael561 Jan 17 '17 at 17:41
• $3^n$ tends to be quite odd ... – Hagen von Eitzen Jan 17 '17 at 17:41
• $3^n-1$ is even for every $n$, and the only even prime is $2$. – barak manos Jan 17 '17 at 17:42
• I'll upvote this for exploring with Mathematica. But having seen the result you guessed, you should have written down the values of $3^2-1$, $3^3-1$ and $3^4-1$, Then you might have noticed that you were gettng even numbers - see @Reese 's answer. – Ethan Bolker Jan 17 '17 at 17:47
• @TheCount: While you are allowed to start counting with one, some of us consider zero a natural number. Zero is not a prime (zero divisors are excluded by definition) – hardmath Jan 17 '17 at 17:51

A good approach to problems of this form is to assume (1) that there are no primes of this sort other than the easy example, and (2) that there's a simple reason, like a single number that must divide all of them. To determine what this "simple reason" is, look at the first few examples: $2, 8, 26, 80, \ldots$. The one factor these all have in common is $2$. So it's reasonable to suppose that every example will be even; we just need to know why. But then the answer is evident: $3^n$ is a product of odd numbers, so it will always be odd. An odd number minus one is always even. And if $n > 1$, then $3^n$ will be greater than $3$, so $3^n - 1$ will always be an even number greater than two - and hence will not be prime.
A little mistake when you say "for $n=0$ is gives you $2$", you want to say "for $n=1$.
$$3^n-1=0\pmod 2$$
(it is always even since $3^n$ is always odd).
So $3^n-1$ can not be a prime once it is greater than $2$. Therefore, the only case acceptable if $3^n-1=2$, which is the case for $n=1$.