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?
$\endgroup$ 93 Answers
$\begingroup$you can write more generally that the product of 2 odd integers is odd
let $a = 2k+1$ and $b = 2p+1$
then $ab = (2k+1)(2p+1) = 2(2kp+k+p)+1 $ which is the general format of odd numbers
$\endgroup$ 3 $\begingroup$It's correct, but I would go a slightly different route. After proving that $n^2 = 4k^2 + 4k + 1$, I would say that $4k^2 + 4k$ is even, since $$\frac{4k^2 + 4k}{4} = k^2 + k$$ (we can even say that $4k^2 + 4k$ is doubly even). And when you add $1$ to an even number, you obtain an odd number.
$\endgroup$ $\begingroup$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 $1^2=1$.
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