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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.

Connor Harris
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Andrew Brooks
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  • Do either of the questions cited actually address this question? In the first, $ad-bc=-4$. In the second, $\gcd(a,b)\mid2$ – robjohn Oct 16 '19 at 18:26
  • @robjohn It's an abstract dupe of the general answers there (viz. determinant = $4$). If I overlooked something then please do elaborate. – Bill Dubuque Oct 19 '19 at 00:47
  • @BillDubuque: abstract duplicate? I'm not sure exactly what that means, but I don't see an answer to those questions that, without a fair amount of extra work, answers this question. Sure, there are starting points and hints, but no answers. I might even go further and show that $$ (a+2b,a-2b)=\left{\begin{array}{} 2&\text{if }a\equiv0\pmod4\ 1&\text{if }a\equiv1\pmod4\ 4&\text{if }a\equiv2\pmod4\ 1&\text{if }a\equiv3\pmod4\ \end{array}\right. $$ – robjohn Oct 19 '19 at 12:15
  • @robjohn Various linked answers show (generally) how to rigorously deduce that the gcd $,d\mid 4,,$ which answers both of the specific questions asked in the final paragraph. What remains is very easy, viz. to show that the values $,d = 1,2,4,$ actually do occur, but this is not what was explicitly asked about (and is probably also a dupe given that we likely have hundreds of such problems by now). – Bill Dubuque Oct 19 '19 at 14:31
  • @Andrew $,d\mid na,nb\iff d\mid (na,nb) = n(a,b) = n\ $ follows by the GCD Universal Property and GCD Distributive Law. These are fundamental GCD laws that one should master as early as possible in order to become proficient at number theory. – Bill Dubuque Oct 19 '19 at 14:42

2 Answers2

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.

Arthur
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1

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 $d=4$

lhf
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