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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

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

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