Prove that if $ a\mid b$ and $ a+b$ is odd, then $a$ is odd

Question

The question is to prove that if $a,b\in\mathbb{Z}$, $\ a\mid b$, and $a+b$ is odd, then $a$ is odd.

I started by considering a direct proof. $$\text{Assume }\ a+b\text{ is odd. Then }a+b=2k+1,\text{ where } k\in\mathbb{Z}.$$ I've considered using the fact that the sum of an even and an odd integer is odd, and the given fact that $a\mid b$, but I've encountered a mental block. Any guidance towards the right direction would be wonderful.

Answer 1

To give an odd sum, the parity must be opposite. So one is odd and the other is even. If $a$ were even it cannot divide odd $b$ (because $b=ka$ would imply even $b$). So $a$ is odd.

Answer 2

Assume $a|b$. If $a$ is even, then $b$ must be even, so $a+b$ must be even.

Therefore also the contrapositive: if $a+b$ is odd, then $a$ is odd.

Answer 3

$a|b$ so $a|a+b$, which is odd, so $a$ must be odd.

Answer 4

Assume $a$ is even;

Since $a+b$ is odd, $b$ is odd.

Then $a \not |$ b$, a contradiction.

Answer 5

Continuing your line of thought.

$$a+b=2k+1-\text{odd} \\ a\mid b\Rightarrow b=am, \ \text{so:}\\ a+am=2k+1 \Rightarrow a=\frac{2k+1}{m+1} -\text{odd, because the numerator is odd}.$$

Answer 6

Since $a + b$ is odd, then exactly one of $a$, $b$ is odd.

Furthermore, since $a|b$, then $a$ cannot be even (for an even integer cannot divide an odd).

Therefore, $a$ must be odd.