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