Proof by contradiction algebra Let n be an integer. Prove that if $n^2+2n-1$ is even, then $n$ is odd.
This is what i have tried so far 
If $n$ is even then $n = 2k$
And $n^2+2n-1$ is odd
$$(2k)^2+2(2k)-1 = 4k^2+4k-1$$

If $n^2 + 2n - 1$ is even, then
$(n + 1)^2 = n^2 + 2n + 1 = (n^2 + 2n - 1) + 2 \tag 1$
is also even; now, the square of an even is even, since
$n = 2k \Longrightarrow n^2 = 4k^2 = 2(2k^2) \Longrightarrow 2 \mid n^2; \tag 2$
likewise, the square of an odd is odd:
$n = 2k + 1 \Longrightarrow n^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1; \tag 3$
it follows that $n + 1$ is even, and hence
$n = (n + 1) - 1 \tag 4$
is odd; this may more formally seen by writing
$n + 1 = 2k$
$\Longrightarrow n = 2k - 1 = (2k -2) + 1 = 2(k - 1) + 1, \; \text{odd}. \tag 5$
$OE\Delta$.
The contradiction in the above proof is rather subtly buried, and not explicitly mentioned; it occurs when we infer $n + 1$ is even from $(n + 1)^2$ is even; there we tacitly assume $n + 1$ is odd which, in light of (2) and (3), forces $(n + 1)^2$ odd, contradicting the demonstrated even-ness of $(n + 1)^2$.
