Re: Romanian math olympiad problem

In article <24018139.1127973621756.JavaMail.jakarta@xxxxxxxxxxxxxxxxxxxxxx>
Hilbert <hilbertspace@xxxxxxxxx> writes:
>so difficult to prove :(
>
>
>Let x and y be real numbers such that the
set { cos ( nx ) + cos ( ny ) | n positive integer } is finite.
Prove that x and y are rational multiples of PI.

Let frac(z) = z - floor(z) denote the fractional part of a real number z.

Given a positive integer m, map it to

f(m) = [frac(m*x/(2*pi)), frac(m*y/(2*pi))]

in the unit square. The image is bounded, so, given epsilon > 0,
we can find positive integers n1 > n2 such that

abs(frac(n1*x/(2*pi)) - frac(n2*x / (2*pi)) < epsilon
abs(frac(n1*y/(2*pi)) - frac(n2*y / (2*pi)) < epsilon

>From this, we can get

cos((n1 - n2)*x) + cos((n1 - n2)*y)

arbitrarily close to 2. Since there are only finitely many
possible values of the sum, it must eventually equal 2,
in which case we're done.

--
The 2nd John Adams president lost to the winner of the Battle of New Orleans.
The 2nd George Bush president lost the Battle of New Orleans.

pmontgom@xxxxxx Microsoft Research and CWI Home: Bellevue, WA
.

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