If $\gcd(a,b)=1$ , then $a-b$ does not divide $a+b$?

Question

I think the following statement is true:

Suppose $a,b\in \mathbb{N}^+$, such that $\gcd(a,b)=1$ and $|a-b|\geq\mathbf3$. Then $a-b$ does not divide $a+b$.

Can you help me to solve this problem?

Answer 1

If positive integer $d$ divides both $a+b, a-b$

$d$ must divide $a+b\pm(a-b)$

$d$ must divide $2(a,b)=2$

So, $a+b,a-b$ can not have a common divisor $>2$

But $|a-b|\ge3>2$

Answer 2

Here is a Bezout Identity approach.

Assume that $a-b\mid a+b$. Then $$ \gcd(a-b,a+b)=a-b $$ Since there are $x,y$ so that $ax+by=1$, we have $$ \begin{align} (a-b)(x-y)+(a+b)(x+y) &=2(ax+by)\\ &=2 \end{align} $$ Thus, $\gcd(a-b,a+b)\mid2$. Therefore, $a-b\mid2$, which contradicts that we are given $|a-b|\ge3$.

Therefore, our assumption that $a-b\mid a+b$ is false.

Answer 3

From here we have if gcd$(a,b)= 1$, then gcd$(a-b,a+b) =1$ or gcd$(a-b,a+b) =2$.

Since $a-b\geq 3$ then gcd$(a-b,a+b) \neq a-b$, then $a-b$ does not divide $a+b$.