How to prove that if c is an odd integer that divides the sum and the difference of two integers a and b, then c divides both a and b? I have two questions regarding this. I'm new to proving so these may seem silly.

*

*Is it better to prove directly or by using its contrapositive? Directly, I could prove by applying a bit of algebra and using its contrapositive, I have to show that c does not divide a or c does not divide b would give c does not divide their sum and difference. I hope this is correct.

*Does c being odd have anything to do with the proof? If yes, how do I show it in the proof? Because when I try proving using c as any integer, I'm getting my answer.

 A: $c$ being odd will be important to the proof. Here is a counter if $c=2$ and $a$, $b$ are odd, then $a+b, a-b$ are even, but $c$ doesn't divided $a$ or $b$.
Here is a direct proof. $c$ divides $a+b$ means that there exist an integer $k_1$ such that $a+b = k_1 c$. Similarly there exists an integer $k_2$ such that $a-b = k_2c$. Adding both equations together, we get that $2a = (k_1+k_2)c$. Now $c$ is odd, however, $2a$ is even, Thus we must have that $k_1+k_2$ is even. Thus $a = \frac{(k_1+k_2)}{2}c$ where $\frac{(k_1+k_2)}{2}$ is an integer.
Now if $c$ hadn't been odd, then we could have had that $k_1+k_2$ was odd, then $\frac{(k_1+k_2)}{2}$ wouldn't have been an integer.
So now $a = k_3c$. Then using $a+b = k_1c$, we see that $b = (k_1-k_3)c$. Thus, $c$ divides $a$ and $b$.
A: I think it is best to prove this directly.
Suppose that $c$ is odd, and that $c \mid (a + b)$ and $c \mid (a - b)$ where both $a$ and $b$ are integers.
Let $A = a + b$ and $B = a - b$.  Then we obtain
$$\Bigg((c \mid A) \text{ and } (c \mid B)\Bigg) \implies \Bigg(c \mid (A + B) = 2a\Bigg).$$
Now, since $c$ is odd, $\gcd(c,2)=1$, which implies that $c \mid a$.
We also obtain, similarly,
$$\Bigg((c \mid A) \text{ and } (c \mid B)\Bigg) \implies \Bigg(c \mid (A - B) = 2b\Bigg).$$
Now, since $c$ is odd, $\gcd(c,2)=1$, which implies that $c \mid b$.

We therefore conclude that $c$ divides both $a$ and $b$.
A: From a linear algebra perspective, the basic reason for this is that $a+b$ and $a-b$ form a basis for the span of $a$ and $b$. If we take $u=a+b$ and $v=a-b$, then $a=\frac{u+v}2$ and $b=\frac{u-v}2$. So we can restate the claim as "If $c$ is an odd integer that divides both $u$ and $v$, then $c$ divides both $\frac{u+v}2$ and $\frac{u-v}2$".
This statement is easy to prove: divisibility is closed under addition and subtraction (that is, a multiple of $c$ plus or minus another multiple of $c$ is also a multiple of $c$). And since $c$ is odd, we don't have to worry about dividing by $2$; if $u+v$ is divisible by $c$, then so is half of $u+v$.
