Odd and even number proofs Let $n$ be an integer. Prove that if $n^2+2n-1$ is even than $n$ is odd.

Proof: $$n^2+2n-1=2n$$ $$n^2-1=0$$ $$(n-1)(n+1)=0$$ $$n=-1,1$$

Which are odd. 
Is this a complete proof? I feel like it only proves $n=-1,1$ not an odd number.
You are making a mistake as follows: Suppose that $n^2+2n-1$ is an even number. This means it is two times some integer. But what you seem to be doing, is equating it to $2n$, which makes it a specific integer, since $n$ is fixed (once we calculate $n^2+2n-1$).

On the other hand, every even integer is given by $2k$, where $k$ is an integer.

To do this question, suppose that $n^2+2n-1 = 2k$, then transposing, $n^2-1 = 2(k-n)$. Hence, $(n-1)(n+1) = 2(k-n)$.

At this point, it is enough to see that if $n$ was even, both $n-1$ and $n+1$ are odd. 

The right hand side is even (divisible by $2$), whereas the left hand side, being a product of odd numbers, is odd. This is a contradiction.

Hence, $n$ must be odd.
Consider:

$EVEN = 2 \cdot ODD$ as you have stated or in more academic way:



$EVEN = 2 \cdot ODD + 0||2$



Can you see the difference?

By the way:



$ODD = 1 \cdot 1 + EVEN$