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?
 A: Suppose $5n^2-3=k$, where $k\in\mathbb{N}$ is even. Then 
$$ 5n^2=k+3$$
and we know that $k+3$ is odd. Next, we have
$$ n^2=\frac{k+3}{5}$$
which must be odd, because any odd divided by an odd is odd. Finally, we have that 
$$n=\sqrt{{\frac{k+3}{5}}}.$$
The square root of any odd perfect square must also be odd, so we know that $n$ is odd. 
A: $$5n^2-3=2k+1$$
$$5n^2=2k+4$$
So $n^2$ must be even. So $n$ is even.
A: This is kinda soft, but you know that $x^2$, if x is odd positive, is always odd. You also know that odd times odd is also odd. And that odd +odd is even. Therefor you can almost deduce the answer.
A: $$5n^2-3\equiv 0 \pmod 2 \iff 5n^2\equiv3\pmod 2 $$
As
$$5\equiv 1, 3 \equiv 1 \pmod 2$$
We have that
$$5n^2\equiv 3 \pmod 2 \iff n^2 \equiv 1 \pmod 2$$
Suppose $n$ is even, that is $n\equiv 0\pmod 2$, then $n^2\equiv 0 \pmod 2$, thus $n$ is odd.
A: For contradiction, assume $n$ can be even. Let $n=2k$. Then $5n^2-3=2\left(10k^2-2\right)+1$ is odd, contradiction.
By contraposition, you can prove analogously that if $n$ is even, then $5n^2-3$ is odd.
A: To prove IF P THEN Q by contradiction you have to prove IF NOT Q and P THEN impossible.
Thus to prove by contradiction:
Assume $5n^2 - 3$ is even and $n$ is even.  If $5n^2 - 3$ is even then $5n^2$ is odd.  If $n$ is odd then $n^2$ is odd.  So $5*n^2 = 5*\text{ an odd number}  = 5n^2 = \text {an even number}$.  That's a contradiction.  So if $5n^2 - 3$ is even th3n $n$ also even is impossible. So n is odd.
One type of proof by contradiction is a proof by contrapositive.  To prove IF P THEN Q.  You can prove IF NOT Q THEN NOT P (thus the only way P can be true is if Q is also.)  (The contradiction is NOT P and P can't both be true.)
A proof by contrapositivve:
Suppose $n$ is even.  Then $5n^2$ is even.  So $5n^2 - 3$ is odd.  So if $5n^2 -3 $ is even, it must be $n$ is odd.
[A third of proof is a direct proof:  If $5n^2 - 3$ is even then $5n^2$ is odd. So $n^2$ is odd. So $n$ is odd.]
What you tried to do was 
IF Q THEN P.  If $n$ is odd then $5n^2 -3$ is even.  In this case that happens to be true but it isn't what was meant to be proven.
Consider IF p is a prime > 2 THEN p is odd.  You can not prove IF p is odd greater than 2 THEN p is prime.  That just isn't true.  And it's not a contradiction nor a contrapositive.
That was what your professor objected to, not that you attempted a proof by contradiction.
As it turns out $n$ is odd $\iff 5n^2 - 3$ is even.  You tried to prove $n$ odd $\implies 5n^2 - 3$ even.  But the exercise was to prove $5n^2 - 3$ even $\implies n$ odd.
A: 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.
A: 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.
