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

We have that $\gcd(pq + r,p) = \gcd(p,r)$ and since $12n - 60 = 3(4n-5)+45$, we have that $\frac{4n-5}{60-12n}$ can't be reduced if and only if $\gcd(4n-5,45) = 1$ and since $45=3^2\cdot 5$, this happens if and only if $3\nmid 4n-5$ and $5\nmid 4n-5$.
Now,
$4n-5\equiv 0  \pmod 3\iff 4n \equiv 8 \pmod 3 \iff n\equiv 2\pmod 3$ (last equivalence is due to $\bar 4$ being invertible in $\mathbb Z/3\mathbb Z$)
$4n-5\equiv 0\pmod 5\iff 4n\equiv 0\pmod 5 \iff n\equiv 0\pmod 5$ (last equivalence is due to $\bar 4$ being invertible in $\mathbb Z/5\mathbb Z$)
Thus, $\frac{4n-5}{60-12n}$ can't be reduced if and only if $n\not\equiv 2\pmod 3$ and $n\not\equiv 0\pmod 5$.
