Prove for every three integers $a$, $b$ and $c$ that an even number of the integers $a + b$, $a + c $and $b + c$ are odd. [duplicate]

Is my proof correct? Asking because it differs from solution in by book. Thanks in advance.

Lemma 1 given: "If the sum of two integers is even then they have the same parity"

Proof by contradiction.
Assume that an odd number is odd.

Case 1: 1 number is odd:
WLOG let $a + b$ be odd. Thenn by lemma 1 $a$ and $b$ have different parity. WLOG let $a$ be even. Since $a$ is even $c$ is also even, since $c$ is even $b$ is also even, which is a contradiction.

Case 2: all three numbers are odd: WLOG let $a$ be even, since $a$ is even $c$ is odd, since $c$ is odd $b$ is even, which results in both $a$ and $b$ being even which is a contradiction.

$\blacksquare$

Solution 1:

Yes, this should be correct. Your application of WLOG looks correct to me. However, I would recommend explaining why you say "since $a$ is even $c$ is also even" and "since $c$ is even $b$ is also even" - I know it's because $a+c$ and $b+c$ are even due to your assumption that $a+b$ is the odd one, but it would make it more thorough to state this.

There will usually be many ways to prove anything so don't worry that it differs from the book. It's a bit more complicated than needed though - if you assume that an odd number of $a+b,b+c,a+c$ are odd, then $(a+b)+(b+c)+(a+c)=2(a+b+c)$ is odd which is clearly impossible - but yours is still a valid proof.