IMO 1961

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Problem A3

Solve the equation cosnx - sinnx = 1, where n is a natural number.

 

Solution

Since cos2x + sin2x = 1, we cannot have solutions with n not 2 and 0 < |cos x|, |sin x| < 1. Nor can we have solutions with n=2, because the sign is wrong. So the only solutions have sin x = 0 or cos x = 0, and these are: x = multiple of π, and n even; x even multiple of π and n odd; x = even multiple of π + 3π/2 and n odd.

 

 

Solutions are also available in:   Samuel L Greitzer, International Mathematical Olympiads 1959-1977, MAA 1978, and in   István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.

 

3rd IMO 1961

© John Scholes
jscholes@kalva.demon.co.uk
19 Sep 1998
Last corrected/updated 24 Sep 2003