The full question states: Let $p$ be a prime. Let $f(x) = 3x+1$ and $g(x) = 6x+1$. Show that: if there exists $x\in \Bbb N$ such that $f(x) = p$, then there exists $y\in \Bbb N$ $g(y) = p$.
My attempt at this was quite simple. I said that as $p$ is prime, and cannot equal 2, as $3x + 1 \not= 2$, $p$ must be odd, therefore $x$ is even. I've got a proof for that as well, but I'm quite confident that that's fine.
Thus, $x$ can be written as $2y$. $$3(2y) + 1 = p$$ $$6y + 1 =p$$ Thus, if $f(x) = p$, then $g(y) =p$.
I'm just not sure if this is a reasonable proof.
