Prove that $5n^2 - 3$ even $\implies n$ odd I tried to prove this by contradiction. I used contradiction to show that if $n$ is odd then $5n^2 - 3$ is even; but my Professor said this is not a correct answer to the question: you need to prove that if $5n^2 - 3$ is even then $n$ is odd. Why is what I said wrong, and how do I fix it?

Can be shown by a parity argument.
The lowest bit defines the oddness or evenness.
From $5n^2 - 3$ we can easily get the parity of the result as $(1.n_0.n_0)\oplus1$ which gives us $n_o\oplus1$
For the OP, $\oplus$ is the exclusive or operation.
Let the least significant bit of $n$ be $n_0$ ( = parity of $n$ ).
As an even $n$ will be $n_0=0$ the expression's parity is $0\oplus1=1$, which is odd.   And of course an odd $n$ will get an even parity.
Can be shown by simple substitution
Another method is simply to try $n = 2k$ and $n = 2k+1$ for an even and odd $n$ respectively.
For $n = 2k$ we get $5n^2-3 = 5(4k^2)-3$ which is always odd.
For $n = 2k+1$ we get $5n^2-3 = 5(4k^2)+5(4k)+5(1)-3$ which is always even.
