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        riddles >> putnam exam (pure math) >> x to the 2005
        
(Message started by: THUDandBLUNDER on Feb 1st, 2005, 9:38am)
Title: x to the 2005
Post by THUDandBLUNDER on Feb 1st, 2005, 9:38am
If x + 1/x = -1 find the value of x2005 + 1/x2005
Title: Re: x to the 2005
Post by Barukh on Feb 1st, 2005, 10:11am
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Rewriting the equation in the form x2 + x + 1 = 0, we see that x is one of the 2 complex cubic roots of unity, while 1/x is the other. WLOG, let x = e2[pi]i/3, then 1/x = e4[pi]i/3.

Referring to the plane of complex numbers, we see that raising x to the n-th power is a rotation of the point (1,0) counter-clockwise by the angle 2[pi]n/3, and analogously for 1/x. When n [ne] 0 mod 3, xn + 1/xn = x + 1/x = -1. When n = 0 mod 3, xn = 1/xn = 1.

Since 3 doesn’t divide 2005, the answer is -1.
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