Can you prove my Prime assumption wrong?

Question

Right, so I was playing about with prime numbers and came across something very odd and interesting. Check this out.

IF $\frac{3x+1}{2^n} \in 2\mathbb{N}-1$

WHERE $n \in \mathbb{N}, x \in 2\mathbb{N}-1$

THEN $\frac{3x+1}{2^n} \in \mathbb{P}$

I can't seem to find a value where this isn't true. And I've tried small numbers like 115 and larger numbers like numbers 57,634,937 and I always landed a Prime. Every time.

Prove me wrong. Not saying I'm right, would be nice to see if there is an exception as I can't seem to find one.

EDIT

Ok, clearly this isn't true, but what if I said...

$\frac{3x+1}{2^n} = r$

SUCH THAT $r = mp$

WHERE $m\in \mathbb{N}, p \in \mathbb{P}, p > 2$

Answer 1

This is false. For instance, if $x=1,$ then $3x+1 = 4$ and $n=2$ gives $4/2^2 = 1$ is not prime.

Answer 2

It's pretty easy to come up with a counterexample here, as I and others have done. Let me explain how we did it.

Your statement is: suppose $x$ is an odd natural, $n$ is a natural number, and $\frac{3x + 1}{2^n}$ is odd. Then $\frac{3x + 1}{2^n}$ is odd.

To come up with a counterexample, we try some other value that $\frac{3x + 1}{2^n}$ could be. We want an odd number which is not prime - let's try $25$ (the smallest odd composite not divisible by 3 - we know the number can't be divisible by 3 since the numerator $3 \cdot x + 1$ is not divisible by 3).

Now clearly, $25 \equiv 1 \mod 3$. So we can write $25 = \frac{3 \cdot 8 + 1}{2^0}$. This doesn't quite work since $x$ isn't odd, but we can try a little harder. We need $25 \cdot 2^n$ to be odd.

So let's try $n = 2$. Then we have $25 = \frac{100}{4}$, and $100 = 33 \cdot 3 + 1$. This gives us a counterexample.