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 enter image description here

  • $\begingroup$ So, what's the method in the Monthly? You wouldn't want us to waste our time just reinventing something ugly. $\endgroup$ – Gerry Myerson Jun 19 '13 at 12:46
  • $\begingroup$ oh,Thank you ,you can find this solution in net $\endgroup$ – math110 Jun 19 '13 at 12:48
  • $\begingroup$ You can find it if you have access to the Monthly, but not everyone does. $\endgroup$ – Gerry Myerson Jun 19 '13 at 12:52
  • $\begingroup$ oh., I find now. $\endgroup$ – math110 Jun 19 '13 at 13:24
  • $\begingroup$ I find it,Thank you,becasuse I have see this problem is china book, $\endgroup$ – math110 Jun 19 '13 at 13:49

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 $\pm1,\pm2,\pm3,\pm4$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.