Use a proof by contraposition to show that if 𝑛𝑛2 + 1 is even, then 𝑛𝑛 is odd.

The first step in a proof by contraposition is to assume that the conclusion of the conditional statement "If $n^2 + 1$ is even, then $n$ is odd" is false; namely, assume that $n$ is even.

Then, by the definition of an even integer, $n = 2k, k\in \Z.$

Substituting $2k$ for $n,$ we find that 

$n^2+1=(2k)^2+1=4k^2+1=2(2k^2)+1$

This tells us that $n^2+1$is odd, and therefore not even. This is the negation of the premise of the theorem.

Because the negation of the conclusion of the conditional statement implies that the hypothesis is false, the original conditional statement is true.

Our proof by contraposition succeeded; we have proved

the theorem " If $n^2 + 1$ is even, then $n$ is odd".