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?

You were asked to prove that if $5n^2-3$ is even, then n is odd. But what you proved was that if n is odd, then $5n^2-3$ is even. 
$A \implies B$ is equivalent to $\neg B \implies \neg A $. So if you prove $\neg B \implies \neg A $, then you've proved $A \implies B$. This is an example of proof by contradiction. Assume the negation of the conclusion. Prove that this leads to a negation of one of your givens.
$A \implies B$ is NOT equivalent to $B\implies A$. For example $(x=2) \implies (x^2=4)$ is true. $(x^2=4) \implies x=2$ is false.
So for your question assume n is even, show that this means $5n^2-3$ is odd. Then you've done the right proof.
Suppose the question asked you to find out if this statement is true or false: "If $2n^2-3$ is odd then n is even". This statement is false because we can have n odd and $2n^2-3$ as odd. But if you assume n is even you will get $2n^2-3$ is odd. So your method would give the wrong result.
