Prove that for all integers $a$ and $b$ that $a + b$ and $a − b$ are either both odd or both even.

Question

Prove that for all integers $a$ and $b$ that $a + b$ and $a − b$ are either both odd or both even.

Stumped on this proof. I've only been able to figure it out assuming that both a and b are even:

$a = 2k$ and $b = 2n$

$2k + 2n = 2(k + n)$, definitely even. $2k - 2n = 2(k - n)$, also definitely even.

I don't think that this is the best way to go about proving this though.

Answer 1

let $a,b,p,q \in \mathbb{Z}$ $$ (a \mbox{ is even}) \land (b \mbox{ is even}) \implies \begin{cases} a=2p\\ b=2q \end{cases} \implies \begin{cases} a+b=2(p+q)&\mbox{(even)}\\ a-b=2(p-q)&\mbox{(even)} \end{cases} $$ $$ (a \mbox{ is even}) \land (b \mbox{ is odd}) \implies \begin{cases} a=2p\\ b=2q+1 \end{cases} \implies \begin{cases} a+b=2(p+q)+1&\mbox{(odd)}\\ a-b=2(p-q)-1&\mbox{(odd)} \end{cases} $$ $$ (a \mbox{ is odd}) \land (b \mbox{ is even}) \implies \begin{cases} a=2p+1\\ b=2q \end{cases} \implies \begin{cases} a+b=2(p+q)+1&\mbox{(odd)}\\ a-b=2(p-q)+1&\mbox{(odd)} \end{cases} $$ $$ (a \mbox{ is odd}) \land (b \mbox{ is odd}) \implies \begin{cases} a=2p+1\\ b=2q+1 \end{cases} \implies \begin{cases} a+b=2(p+q+1)&\mbox{(even)}\\ a-b=2(p-q)&\mbox{(even)} \end{cases} $$ or more elegantly: $$ (a+b)-(a-b)=2b \iff (a+b)=(a-b) + 2b \implies (a+b)=(a-b) + \mbox{(even number)} $$ so considering $a-b$ parity and using the above result: $$ (a-b) \mbox{ is odd} \implies (a+b)=\mbox{(odd)} + \mbox{(even)}=\mbox{(odd number)}\\ (a-b) \mbox{ is even} \implies (a+b)=\mbox{(even)} + \mbox{(even)}=\mbox{(even number)} $$

Answer 2

Case $1$: $a$ is odd , $b$ is odd. Then $a=2m+ 1,b=2n+ 1$. So $a+b=2(m + n+1) $ $and $ $a-b=2(m-n-1),$ both of which are even. Similarly check for $a=2m-1$ and $b=2n-1$

Case $2$: $a$ is even , $b$ is odd. Then $a=2m,b=2n+ 1$. So $a+b=2(m + n)+1 $ $and $ $a-b=2(m-n)-1$ both of which are odd. Similarly check for $b=2n-1$

Case $3$: $a$ is odd , $b$ is even. Then $a=2m+1,b=2n$. So $a+b=2(m + n)+1 $ $and $ $a-b=2(m-n)+1$ both of which are odd. Similarly check for $a=2m-1$

Case $4$: $a$ is even , $b$ is even. Then $a=2m,b=2n$. So $a+b=2(m + n) $ $and $ $a-b=2(m-n)$ both of which are even. $\blacksquare$

Answer 3

Hint: Work with parity of $a$ and $b$ (i.e. mod 2). Take the 4 cases (Odd,odd), (odd, even) etc and conclude the result.

Answer 4

Without having to look at cases:

Let $d=(a+b)-(a-b)$, think of it as the distance between $a+b$ and $a-b$.

Then, clearly, $d=2b$. Since the distance between the two numbers is even, both numbers must have the same parity (either both even or both odd).