<s>[INST] Find all the numbers $n$ such that $\frac{4n-5}{60-12n}$ can't be reduced. 
Find all the numbers $n$ such that $\frac{4n-5}{60-12n}$ can't be reduced.

Attempt:
$$\gcd(4n-5,60-12n)=(4n-5,-8n+55)=(4n-5,-4n+50)=(4n-5,45)$$
$$n=1: (4-5,45)=1\quad \checkmark\\
n=2: (3,45)=3\quad \times\\
n=3: (7,45)=1\quad \checkmark\\
n=4: (11,45)=1\quad \checkmark\\
n=5: (15,45)=15\quad \times\\
n=6: (19,45)=1\quad \checkmark\\
n=7: (23,45)=1\quad \checkmark\\
n=8: (27,45)=9\quad \times\\
\vdots$$
So the answer is that it can't be reduced for $n=1,3,4,6,7,..$ i.e
$$\bigg\{n\bigg|n\notin \begin{cases}a_1=2\\a_n=a_{n-1}+3\end{cases}\bigg\}$$
I want to verify that my solution is correct
 [/INST] 
I want to verify that my solution is correct

Your solution is not correct.
For $m\not=1$,
$$\frac{4\times\color{red}{5m}-5}{60-12\times\color{red}{5m}}=\frac{5(4m-1)}{5(12-12m)}$$
can be reduced.

If $n$ is of the form $3k+2$, then
$$\frac{4(3k+2)-5}{60-12(3k+2)}=\frac{3(4k+1)}{3(-12k+12)}$$
can be reduced.
If $n$ is not of the form $3k+2$, then since $4n-5$ is not divisible by $3$, we get
$$\gcd(4n-5,12(5-n))=\gcd(4n-5,4(5-n))=\gcd (4n-5,15)=\gcd(4n-5,5)$$
Now note that
$$\gcd(4n-5,5)=1\iff n\not\equiv 0\pmod 5$$
Hence, the answer is that $$n\not\equiv 2\pmod 3\quad\text{and}\quad n\not\equiv 0\pmod 5.$$
</s>