2

Find all the numbers $a,b$ such that $\frac{2a-b}{2a+b}$ can't be reduced

Attempt:

For $a:$

$$\gcd(2a-b,2a+b)\\ =\gcd(2a-b,4a)\\ \boxed{\{a:4a\nmid2a-b\}}$$

For $b:$

$$\gcd(2a-b,2a+b)\\ =\gcd(2a-b,-2b)\\ \boxed{\{b:-2b\nmid2a-b\}}$$

I have a bad feeling about my answer

Related

Community
  • 1
Error 404
  • 1,892
  • 1
  • 11
  • 23
  • I am just asking : Can we write gcd$(a,b)=-$gcd$(-a,b)$? – Qwerty Jul 06 '16 at 14:56
  • 1
    @Qwerty I don't think so, I think that $\gcd(a,b)>0$ for all $a,b\in \mathbb Z$ – Error 404 Jul 06 '16 at 15:03
  • Okay.I got it . – Qwerty Jul 06 '16 at 15:07
  • 1
    @Qwerty As Error 404 says, $\gcd$ picks out the greatest positive common divisor (equivalently the greatest integer divisor) (or 0 in the case of $\gcd(0,0)$. Morally speaking it should perhaps return an ideal, but it is standard to think of ideals in $\mathbb{Z}$ as corresponding to the nonnegative integer generating them. – Caleb Stanford Jul 06 '16 at 15:08

2 Answers2

1

If $b$ is even, $2|(2a-b,2a+b)$

So, $b$ must be odd.

Now, if integer $d>0$ divides both $2a-b, 2a+b$

$d$ must divide $2a+b\pm(2a-b)=4a,2b$

So, $d$ must divide $(4a,2b)=2(2a,b)$

As $b$ is odd, $d$ must be odd & must divide $(2a,b)=(a,b)$ as $b$ is odd

So, if $(a,b)=1, d=1$

Can you take it from here?

lab bhattacharjee
  • 274,582
1

There is an error in your attempts when you write, e.g. $$ \quad \quad\gcd(2a-b,4a) \\ \to \quad \{a:4a\nmid2a-b\} $$ because it is possible that $\gcd(2a-b,4a)$ is greater than $1$, so that the fraction reduces, even while simultaneously we do not have such a strong statement as $4a$ dividing $2a - b$. (e.g. $2a-b$ is even but not divisible by $4$ or by $a$.)

Instead, here's a useful fact: $$ \gcd(xy,z) = 1 \quad\iff\quad \gcd(x,z) = 1 \text{ and } \gcd(y,z) = 1 $$ Therefore, we have that \begin{align*} \gcd(2a-b,4a) = 1 &\iff \gcd(2a-b,4) = 1 \text{ and } \gcd(2a-b,a) = 1 \\ &\iff b \text{ is odd,} \text{ and } \gcd(a,b) = 1. \\ \end{align*}

This gives you the condition you want: $a,b$ relatively prime and $b$ odd.

Caleb Stanford
  • 45,895