The square of any odd integer is odd. Suppose that $n$ is an odd integer. Then $n = 2k + 1$ for some integer $k$. 
Hence $n^2=(2k+1)(2k+1)=4k^2+4k+1=2(2k^2+2k)+1$.
Since $k$ is an integer, $2k^2 + 2k$ is an integer. Thus $n^2 = 2k' + 1$ for some integer $k'$.
Therefore $n^2$ is odd.
Is this correct?

Using modular arithmetic:
$$n\equiv1\mod2\implies n^2\equiv1\mod2$$ because $n^2\bmod2=(n\bmod2)^2$ and $1^2=1$.

You can even handle the case of even and odd at the same time with
$$n\bmod2=(n\bmod2)^2=n^2\bmod2$$ because $0^2=0$ and $1^2=1$.
