Proof by contradiction involving positive integer prime numbers

Question

CONTEXT: Challenge question set by uni lecturer for discrete mathematics course

Question: Prove the following statement is true using proof by contradiction:

For all positive integers $x$, if $x$, $x+2$ and $x+4$ are all prime, then $x=3$.

I know I'd do this by trying to prove the negation of the statement, but then failing to do so and hence 'contradicting' myself.

I've also found the negation to be that there exists a positive prime integer $x$ such that +2 and +4 are prime, but $x≠3$

I'm stuck at the part of the proof where you show that the negation is false.

Answer 1

Hint: One of $x, x+2, x+4$ must be divisible by $3$.

Answer 2

Assume not, so $\exists x$ positive integer s.t. $x, x+2, x+4$ prime but $x\neq3$.

But we can look at the numbers modulo 3, and realize that there are 3 possiblilities:

1) $x\equiv 0 (mod 3)$. So $x+2\equiv 2 (mod 3)$ and $x+4\equiv 1 (mod 3)$

2) $x\equiv 1 (mod 3)$. So $x+2\equiv 0 (mod 3)$ and $x+4\equiv 2 (mod 3)$

3) $x\equiv 2 (mod 3)$. So $x+2\equiv 1 (mod 3)$ and $x+4\equiv 0 (mod 3)$

In any event, one of the 3 numbers must be divisble by 3, which means it is not prime (unless it is equal to 3). Which is only possible if $x=3$