### 29th Putnam 1968

Problem B3

Given that a 60o angle cannot be trisected with ruler and compass, prove that a 120o/n angle cannot be trisected with ruler and compass for n = 1, 2, 3, ... .

Solution

This is a trap.

The obvious answer is that if we could trisect an angle 120o/n, that would give us an angle 40o/n and hence 60o/3. But we are told we cannot trisect 60o.

This does not work for a slightly subtle reason. We are given the angle 120o/n. That in itself may help us to trisect 60o. For example, it certainly allows us to trisect 60o if n = 6.

So we have to use the standard Galois theory approach. It is not hard to show that a straight-edge allows us to get rational distances and compasses allow us to get any finite number of square roots. So we can construct distances which are in an extension K of the rationals of degree 2n for some n. Constructing an angle is equivalent to constructing its cosine.

So if we are given an angle A, then we can construct distances which are in any extension K of degree 2nover Q(cos A). Now if K is an extension of L of degree r and L is an extension of F of degree s, then K is an extension of degree rs of F. So suppose Q(cos A) has degree u and Q(cos(A/3 ) has degree v. If we can construct A/3 given A, then cos(A/3) must lie in an extension K of Q(cos A) of degree 2n. But K is also an extension of Q(cos(A/3) ) so v/u must be a power of 2.

Finally, it is well-known that Q(cos(360/m) ) has degree φ(m) over Q, where φ is Euler's function (so that φ(m) is the number of 1, 2, ... m-1 relatively prime to m). So we have to show that φ(9n)/φ(3n) is not a power of 2. But φ(m) = m ∏(1 - 1/p), where the product is taken over all primes p dividing m. But 3n and 9n have the same prime divisors, so φ(9n) = 3 φ(3n), and 3 is not a power of 2.