[Math] Direct proof that if mn is odd then m is odd and n is odd

discrete mathematicsfake-proofs

I found the converse here, although that's not what I want.

I have thought of a proof by contradiction and by contraposition, although I can't seem to figure out a way to finish a direct proof.

$mn = 2a + 1
$

If $a = 2kj + (k+j)$ for integers $k$ and $j$, which I got out of my crystal ball, then $mn = 4kj + 2(k+j) + 1$ and $mn = (2k + 1)(2j + 1)$, but then I have to prove that it's possible to write any integer $a$ as $2kj + (k + j)$, which I don't know how.

Any help would be appreciated.

Best Answer

Write $m=2k+a$, where $a=0$ or $1$.

Write $n=2l+b$, where $b=0$ or $1$.

$mn=(2k+a)(2l+b)=4kl+2kb+2al+ab$

Since $mn$ is odd, $ab=1$, which means $a=1$ and $b=1$.