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lemon14
Nov 13, 2010, 09:16 AM
x_1, x_2 are the solution for x^2+x+1=0

How much is n \epsilon N so that (x_1^2+1)^n+(x_2^2+1)^n=-1 to be true?

galactus
Nov 13, 2010, 09:49 AM
The solutions to the quadratic are

x_{1}=\frac{-1}{2}+\frac{\sqrt{3}}{2}i, \;\ x_{2}=-\frac{1}{2}-\frac{\sqrt{3}}{2}i

Sub these into (x_{1}^{2}+1)^{n}+(x_{2}^{2}+1)^{n} and it whittles down to

2cos(\frac{n\pi}{3})

Now, set this equal to -1 and solve for n.