Number of pairs $(m,n)$ of positive integers such that there exists a real number $x$ satisfying $ \sin(mx) + \sin(nx) = 2.$ Find the number of pairs $(m,n)$ of positive integers
with $1\le m<n\le 30$ such that there exists a real number $x$ satisfying
$ \sin(mx) + \sin(nx) = 2.$
This AIME 2021 P7
The range of $\sin x=[-1,1].$
So $\sin(mx)=1=\sin(nx).$
So $mx=90+360k= ,nx=90+360l$
So $x(m-n)=360k-360l=360(k-l).$ We have $(m-n)\le 30$
 A: Hint: Assuming the angles are measured in degrees (note the same result occurs even if they're in radians instead), then using your $2$ equations for $mx$ and $nx$, multiply the first one by $\frac{n}{90}$ and subtract the second one multiplied by $\frac{n}{90}$ to get
$$0 = n - m + 4(kn - ml) \iff n - m = 4(ml - kn) \tag{1}\label{eq1A}$$
Because $n - m$ and $ml - kn$ are both integers, $n - m$ must be an integral multiple of $4$.
Update: Note this is necessary, but it's not always sufficient. For example, with $n = 12$ and $m = 4$, then $n - m = 8 = 2^3$, while $4(ml - kn) = 4(4)(l - 3k) = 2^4(l - 3k)$. Since the right side in \eqref{eq1A} has more factors of $2$ than the left side, it shows the equation is not true in this case.
A: We can take the cases like
1:n-m=4 then n and m should be odd
2:n-m=8 then n and m should not bhle the multiple of 4
3:n-m=12 then n and m should be odd
4:n-m=16 then n and m should not be the multiple of 8
5:n-m=20 n and m should be odd
6:n-m=24 n and m should not be multiple of 4
7:n-m=28 n and m should be odd
It is bit lengthy if anyone has short solution plz tell
