Show that if $3n + 5$ is even, then $n^2$ is odd.

Question

What approach should I take to prove this?

Answer 1

If $3n+5$ is even, then $3n$ is odd (since odd+odd=even, even+odd=odd, even+even=even). Then, $n$ must be odd, since odd*odd = odd, even*odd=even, even*even=even. Since odd*odd=odd, $n^2$ is also odd.

Answer 2

A very quick proof if you know modular arithmetic.

$3n \equiv -5 \equiv - 3 \pmod 2$

$n \equiv -1 \pmod 2$ (division permitted because $(3,2) = 1$)

$n^2 \equiv 1 \pmod 2$ (QED)