Here's the question you clicked on:
JoãoVitorMC
Prove that for any integer n, at least one of the integers n, n+2, n+4 is divisible by 3 .
I don't recall how to organize a proof of this type. But you might take it one case at a time... If n is divisible by 3, then you're done. If n is not divisible by 3, then it must either be 1 greater or 2 greater than a number that is divisible by 3 (i.e., if it was 3 greater than a number divisible by 3, it would also be divisible by 3). If n is one greater, then n + 2 is three greater than the number divisible by 3, therefore n+2 is also divisible by 3. If n is two greater than a number divisible by 3, then n + 4 will make it 6 greater than that number, making n + 4 divisible by 3.
That's not a proof, but maybe you could use that logic to formalize in a proof format.
you could use another variable to help. If n is not divisible by 3, then if m is divisible by 3, n must be m + 1 or m + 2.
so n + 2 = (m + 1) + 2 = m + 3... and since m is divisible by 3, so is m+3
is this make sense? n+(n+2)+(n+4) = 3n+6 , but (3n+6)/3 = n+2 is that sufficient proof?