Question

Q. If (a,b)=1 then prove that
(a-b, a+b, ab)=1

Answer 1

Answer on Question #52569 – Math, Combinatorics – Number Theory

Question. If $(a,b)=1$, then prove that $(a-b,a+b,ab)=1$.

Solution. Notice that the statement $(a,b)=1$ is equivalent to the assumption that there exist integer numbers $p,q$ such that

$ap+bq=1.$

We should prove that $(a-b,a+b,ab)=1$, i.e. that the greatest common divisor of all three numbers $a+b$ $a-b$ and $ab$ is $1$. However, it may happens that some of them are not relatively prime.

For instance $(5,3)=1$, but $(5+3,5-3)=2$. Nevertheless in this case we have that $(5+3,5-3,5\cdot 3)=1$.

It suffices to show that

$(a+b,ab)=1,$

that is $a+b$ and $ab$ has no nontrivial common divisors. This will imply that all three numbers $a-b$, $a+b$ and $ab$ has no nontrivial common divisors, i.e. $(a-b,a+b,ab)=1$. For the proof we need two lemmas.

Lemma 1. If $(a,b)=1$, then $(a,a+b)=(b,a+b)=1$ as well.

Proof. It suffices only to prove that $(a,a+b)=1$. From $ap+bq=1$, we geet

$1=(a,b)=ap+bq=ap-aq+aq+bq=a(p-q)+(a+b)q=(a,a+b)$

Lemma 1 is proved. $\square$

Lemma 2. If $(x,a)=(x,b)=1$ and $(a,b)=1$, then $(x,ab)=1$ as well.

Proof. By assumption there are integer numbers $\alpha,\beta,\gamma,\delta$ such that

$(x,a)$ $=\alpha x+\beta a=1,$

$(x,b)$ $=\gamma x+\delta b=1.$

Multiplying the first equation by $bq$ and the second by $ap$ and adding them we get

$(\alpha x+\beta a)bq+(\gamma x+\delta b)ap=bq+ap.$

Taking to account that $(a,b)=ap+bq=1$, we see that the right hand side is $1$, whence

$(\alpha x+\beta a)bq+(\gamma x+\delta b)ap=1,$

$x(\alpha bq+\gamma qp)+ab(\beta q+\delta p)=1.$

thus $(x,ab)=1$ as well. Lemma 2 is proved. $\square$

Corollary 3. $(a+b,ab)=1$.

Proof. By Lemma 1 we have that $(a+b,a)=(a+b,b)=1$. Then from Lemma 2 applied to $x=a+b$ it follows that $(a+b,ab)=1$. Corollary 3 is proved. $\square$

Now we can complete the proof. Notice that by definition $(a-b,a+b,ab)$ divides $a+b$ and $ab$, whence it must divide $(a+b,ab)=1$. Therefore $(a-b,a+b,ab)=(a+b,ab)=1$.