Let $a, b, c ∈ Z$. Then $ab + ac + bc$ is even if and only if at most one of $a, b$ and $c$ is odd. Is my proof correct? if so, is there something which isn't done nice? I am currently learning proofs on my own and the solutions in my book aren't always complete.
Lemma 1 given: Sum of an odd number of odd numbers is odd.
Lemma 2 given: A product of two integers is odd if both integers are odd.
Proving first implication: $ab+ac+bc$ is even $\Rightarrow$ at most one of $a,b,c$ is odd.
Proof by contradiction:
Suppose that $ab+ac+bc=2k$ and at least 2 of $a,b,c$ are odd.
Case 1: 2 numbers are odd.
WLOG assume that $a$ and $b$ are odd, write $a$ as $2k+1$, $b$ as $2l+1$ and $c$ as $2m$.
$$ab+ac+bc = (2k+1)(2l+1) + (2k+1)(2m) + (2l+1)(2m)$$
$$=(4kl + 2k + 2l + 1) + (4km + 2m) + (4lm + 2m)$$
$$=4kl + 2k + 2l + 4km + 2m + 4lm + 2m + 1$$
$$=2(2kl + k + l + 2km + m + 2lm + m) + 1$$
Since $2kl+k+l+2km+m+2lm+m$ is an integer $ab+ac+bc$ is odd which is a contradiction.
Case 2: three numbers are odd: is a contradiction if we proceed like in case 1.
Proving the second implication: If at most one of $a,b,c,$  is odd $\Rightarrow$ $ab +a c + bc$ is even.
Proof by contradiction:
Suppose that at most one of $a,b,c$ is odd and $ab + ac + bc$ is odd. Since $ab + ac + bc$ is odd,by lemma 1 one or three of the products are odd. By lemma 2 at least two numbers of $a,b,c$ are odd which is a contradiction.
$\blacksquare$
Edit:
Forgot to add second case.

We'll keep in mind that the sum should be even. We'll also use the fact that $ab + bc + ac = b(a + c) + ac$. This will be for the first implication.
If both $a$ and $c$ are even, then $ac$ must be even and $a + c$ must be even. Having an even sum implies that $b(a + c)$ is even and $b$ can be odd or even as desired.
If both $a$ and $c$ are odd, then $ac$ will be odd and $a + c$ will be even. Having an even sum implies that $b(a + c)$ is odd and regardless of the parity of $b$, $b(a + c)$ will be even. A contradiction.
Therefore, the even sum $ab + bc + ac$ implies that, at most one of $a$, $b$, and $c$ is odd.
