# 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.
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.
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.

• For the first part, the argumentation resembles a contrapositive argument, it would be more professional if you pick that way. (Even though this is fine.) The second part is okay, but again you may choose a direct approach. Assuming w.l.o.g that $a$ is the only odd integer, $ab$ and $ac$ are even, since $bc$ is also even, their sum adds up to an even integer. Commented Nov 14, 2021 at 16:39

Your argument is correct. Sometimes though people frown upon proofs by contradiction if you can do it without argument by contradiction.

For example, here you can say: If two or three of the numbers are even, each product is even and therefore the sum is even. If exactly one number is even, then two products are even and one is odd, so the sum is odd. If no numbers are even, each product is odd and therefore the sum is odd.

Still, what you did was fine.

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.

• If $a$ and $c$ are even, $ac$ must be odd is not true. Commented Nov 15, 2021 at 0:01
• Thank you for noticing. I mixed the odd and even case multiple times when I made it. Commented Nov 15, 2021 at 0:03