Let me preface this by saying I am horrible at math, and I apologize for the dumb question.
So, I'm trying to prove that "for all integers 𝑛, if (𝑛^2) + 2 is even, then 𝑛 is even.", and it has to be by contraposition. This is what I have so far:
By contrapositive, this statement is the same as: for all integers n, if n is odd, then (n^2) + 2 is odd
By definition of odd, n = 2k+1 for any integer k
Thus by substitution, ((2k+1)^2) + 2
= 4k^2 + 4k + 3
=4(k^2 + k) + 3 by basic algebra
Obviously I'm trying to get it to take the 2n+1 form for an odd integer, but right now it's stuck at 4n+3. How do I get it into the proper form to complete the proof? Am I missing something obvious?