My apologies. I do not mean that N+2 is congruent to 1(mod3); I mean that if you suppose that N is congruent to 1(mod3) (and thus is of the form 3k+1), then N+2 will be congruent to 0(mod3), which would make your second odd integer not prime. The same reasoning follows if N is congruent to 2(mod3), N+4 will fail to be prime.
These are the three cases Undefined is speaking of.
You do not have to do all that though. What Undefined (I believe) is trying to convey, is that your starting odd integer has to belong to one of three classes: 0, 1 or 2(mod3). It can not belong to 0, as that would make it non-prime. If it belongs to 1(mod3), then one of your other integers will fail (the second one). If it belongs to 2(mod3) then the other integer will fail (the third one). Your then done, as you've shown that all cases result in a list that includes at least one non-prime.That is why I said S is equiv and then choose the appropriate element of S.
Oh, I was reading "sub in" loosely as in,
first we had n to the left of the congruence sign
then we had n + 2 to the left of the congruence sign
so n + 2 was "subbed into" the place holder that is on the left of the congruence sign.
I certainly did not mean to say that $\displaystyle n \equiv n+2\ (\text{mod}\ 3)$. (Which can never be true.)