Let $a$ and $b$ be relatively prime integers. Find all possible values of $\gcd(a+2b,a-2b)$

Question

My attempt: it is known that $(a,b)=1$. Let $d = (a+2b,a-2b)$.

Then it follows that $d|{(a+2b)}$ and $d|{(a-2b)}$, and thus $d|{(m(a+2b)+n(a-2b))}$; that is, $d$ divides all linear combinations of $(a+2b)$ and $(a-2b)$.

If we write $X = a + 2b$ and $Y = a - 2b$, then:

• $2X+2Y=4a$ and thus $d|4a$
• $X-Y=4b$ and thus $d|4b$

Therefore $(4a,4b)=d$ and $(4a,4b)=4(a,b)=4$.

It follows that $(a+2b,a-2b) \leq 4$ and so 4,3,2, and 1 are possible values.

$(a+2b,a-2b) \neq 3$ given it is not a divisor of 4.

Therefore $(a+2b,a-2b) = \{4,2,1\}$

As one could imagine I'm posting because I am not sure of my proof. In particular I'm not certain whether the less than or equal to 4 actually follows from the previous line. As well as this, I'm not really certain my explanation for 3 not being in the answer set is adequate.

Answer 1

Everything before "Therefore $(4a, 4b) = d$" looks good.

From $d\mid 4a$ and $d\mid 4b$, you do not know that $(4a, 4b) = d$. You only know that $d\mid (4a, 4b)$, which is to say, $d\mid 4$.

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

In this case, I think the best option is to actually check. If $a = 6, b = 1$, then we get $d = 4$. If $a = 5, b = 1$, we get $d = 1$. Finally, if $a = 4, b = 1$, we get $d = 2$. So they are indeed all possible.

Answer 2

Here is a different take:

Let $u=a+2b$ and $v=a-2b$. Then $$ \pmatrix{ u \\ v} = \pmatrix{ 1 & \hphantom-2 \\ 1 & -2} \pmatrix{ a \\ b} $$ The matrix above has determinant $-4$ and so, by Cramer's rule, $4 (a \mathbb Z + b \mathbb Z) \subseteq u\mathbb Z + v \mathbb Z$.

We have $a \mathbb Z + b \mathbb Z = \mathbb Z$ because $\gcd(a,b)=1$. We also have $u\mathbb Z + v \mathbb Z = d \mathbb Z$, where $d=\gcd(u,v)$.

Therefore, $4 \mathbb Z \subseteq d \mathbb Z$ and so $d$ divides $4$, that is, $d \in \{1,2,4\}$.

We still need to show that all possible values of $d$ actually occur:

• $a=1, b=0$ gives $d=1$

• $a=0, b=1$ gives $d=2$

• $a=2, b=1$ gives $d=4$