Answer to Question #96099 in Combinatorics | Number Theory for Dina

Question

Let a and b be integers. Prove that (a, b) = 1 if and only if (a + b, ab) = 1.

Answer 1

1)Suppose that $(a,b)=d>1$ . So $a$ and $b$ are divisible by $d$. Then $a+b$ and $ab$ are divisible by $d$, that is $(a+b,ab)$ is divisible by $d$ .

So if $(a+b,ab)=1$, then $(a,b)=1$.

2)Suppose that $(a+b,ab)$ is divisible by some prime $p$. Then $ab$ and $a+b$ are divisible by $p$.

Since $a+b$ is divisible by $p$, there is such $k$ that $a+b=pk$.

Since $ab$ is divisible by $p$, we have that $a$ is divisible by $p$ or $b$ is divisible by $p$.

If $a$ is divisible by $p$, then $a=pl$ for some $l$, then $b=pk-a=p(k-l)$, that is $b$ is divisible by $p$.

By symmetry we obtain that if $b$ is divisible by $p$, then $a$ is divisible by $p$.

In both cases $a$ and $b$ are divisible by $p$, so $(a,b)$ is divisible by $p$.

Therefore if $(a,b)=1$, then $(a+b,ab)=1$.

From 1) and 2) we obtain that $(a+b,ab)=1$ if and only if $(a,b)=1$.