Prove that there are no integers $\csc {\frac{j\pi}{n}}-\csc{\frac{k\pi}{n}}=2$ Prove that there are no integers $j,k,n$ with odd $n$ satisfying
$$\csc {\dfrac{j\pi}{n}}-\csc{\dfrac{k\pi}{n}}=2$$
This problem from $AMM,1999,10630$, but this solution is very ugly,and it's not nice.
I think this problem have other methods,Thank you everyone.
I find this 

 A: For what it's worth, here's what I sent the Monthly in 1999. 
$\def\e{{\bf E}}
\def\q{{\bf Q}}
\def\k{{\bf K}}
\def\f{{\bf F}}
\def\s{\sigma}
\def\t{\tau}
\def\r{\rightline}$
We prove a bit more, namely, that if $\csc(j\pi/n)-\csc(k\pi/n)$ is a 
non-zero rational number and $j$, $k$, and $n$ have no common factor 
then $n=2$ or $n=6$. 
Let $\e_1=\q(\csc{j\pi\over n})$, $\e_2=\q(\csc{k\pi\over n})$, let $\k$ 
be the compositum and $\f$ the intersection of the two fields. Note 
that all the fields are subfields of $\q(e^{\pi i/2n})$, hence they 
are all normal. 
Suppose $[\e_1:\f]>1$. Then there is a non-identity automorphism $\s$ 
of $\e_1$ fixing $\f$. We can extend this to an automorphism $\t$ of $\k$ 
fixing $\e_2$. Apply $\t$ to 
$$
\csc(j\pi/n)-\csc(k\pi/n)=p/q; 
$$ 
it moves 
only the first term in this equation, which is impossible. Thus, $\e_1=\f$. 
The same argument shows $\e_2=\f$, so $\e_1=\e_2$, so $\e_1=\e_2=\k$. Now 
assume $[\k:\q]>1$. Then there is an automorphism $\s$ of $\k$ fixing $\q$ 
such that 
$$
\s\bigl(\csc(j\pi/n)\bigr)=\csc(k\pi/n).
$$ 
Let $\s$ have order $r>1$. 
Then applying $1,\s,\dots,\s^{r-1}$ to $\csc(j\pi/n)-\csc(k\pi/n)=p/q$ 
and summing we get $0=rp/q$, contradiction. 
Thus, $\csc(j\pi/n)$ and $\csc(k\pi/n)$ are rational. This only happens 
when $j/n$ and $k/n$ are of the form $m\pm{1\over6}$ or $m+{1\over2}$ 
for some integer $m$, and we are done. 
The non-zero rational numbers that can occur are $\pm1,\pm2,\pm3,\pm4$. 
