Answer to Question #96782 in Combinatorics | Number Theory for Dina Patel

Question

Let 'a' and 'b' be relatively prime integers. Find all values of (a + 2b, a − 2b).

Answer 1

Given that $a$ and $b$ are relatively prime integers. Then $gcd(a,\ b)=1$ or in equivalent notation $(a,\ b)=1.$

Let $d=(a+2b, \ a-2b).$ Then it follows that $d|(a+2b)$ and $d|(a-2b).$

That is $d|(m(a+2b)+n(a-2b)),$

$2(a+2b)+2(a-2b)=4a$ and thus $d|4a,$

$2(a+2b)-2(a-2b)=4b$ and thus $d|4b.$

Hence $d|(4a,\ 4b)$. We have that $d|4.$

We can rule out every possibility except $d=1,2$ or $d=4.$ However, we do not yet know which of those three values are actually possible.

Let's show that all possible values of $d\in \lbrace 1,2,4\rbrace$ actually occur:

$a=1,b=0$ gives $d=1,$

$a=0, b=1$ gives $d=2,$

$a=2, b=1$ gives $d=4.$

Therefore, the answer is $d\in \lbrace 1,2,4\rbrace.$