Answer
By simple algebraic manipulation, we see that $n^{2}-n+3$$=n(n-1)+3=n(n-1)+2+1$. We now examine two cases: $n$ is even, then $n$ is odd.
Let $n$ be even. By the definition of even, $n=2k$ for some integer $k$. Therefore, we have $n^{2}-n+3=n(n-1)+2+1$$=2k(2k-1)+2+1$$=2[k(2k-1)+1]+1$. This is odd because it is of the form $2x+1$ for some integer $x$, so $n^{2}-n+3$ is odd.
Now let $n$ be odd. By the definition of odd, $n=2k+1$ for some integer $k$. Therefore, we have $n^{2}-n+3=n(n-1)+2+1$$=(2k+1)[(2k+1)-1]+2+1=(2k+1)(2k)+2+1$$=2[k(2k+1)+1]+1$. This is also odd, since it is of the form $2x+1$ for some integer $x$, and so $n^{2}-n+3$ is again odd.
Since $n^{2}-n+3$ is odd in either case, since these are the only two possible cases, and since $n$ was arbitrarily chosen, we conclude that $n^{2}-n+3$ is odd for all integers $n$.
Work Step by Step
In both cases, the assertion that $x$ (representing $k(2k-1)+1$ and $k(2k+1)+1$, respectively) is an integer may be justified by the fact that it is the result of adding, subtracting, and multiplying integers.
