How to prove if $m$, $m + 2$, $m + 4$ are all primes, then $m = 3$

Question

I feel as though I have to use mods, but I'm not sure how exactly to go about this one.

Answer 1

You don't have to use mods but they do make things much simpler.

Suppose $m$ is a positive prime other than 3. Then $m \equiv 1$ or $2 \pmod 3$. In the former case, $m + 2 \equiv 0 \pmod 3$ so it's clearly composite. In the latter case, $m + 4 \equiv 0 \pmod 3$, so again we have a composite.

This leaves $m \equiv 0 \pmod 3$ as the only viable possibility, but then $m$ has to be 3 in order to be prime.

Alternatively you can do things "algebraically": instead of $m \equiv 1 \pmod 3$ you say $m = 3k + 1$, for example.

Answer 2

You can show that one of them must be divisible by $3$. If it is not $3$, it is not prime.

Answer 3

Hint:

$$m(m+2)(m+4)=m^3+6m^2+8m=6m^2+9m+m^3-m\\=3(2m^2+3m)+(m-1)m(m+1)$$

Now among any three consecutive integers, one is divisible by $3$...

Answer 4

One of the three numbers $n$, $n+2$, $n+4$ has to be a multiple of $3$. And only one prime is a multiple of $3$.