# Solve the following equation for x :

$$(x + 1)^{63} + (x + 1)^{62}(x−1) + (x + 1)^{61}(x−1)^{2} + . . .+(x−1)^{63}= 0$$

My approach was:

Its a GP with

$$r= \frac{(x-1)}{(x+1)}$$

Then with the expression:

$ar$$n-1$$=(x-1)$$63 Plugging value of r and a it results to it: n=64 Plugging in GP sum formula i finally get to this : (x-1)$$64$ - $(x+1)$$64 =0 Then what should be done Solve for all 64 values? ............ • this is some moscow olympiad or something – Pole_Star Feb 13 '17 at 13:55 • @dp1611 got it right. – Amar30657 Feb 13 '17 at 13:55 ## 4 Answers Note that we have$$(x+1)^{64}=(x-1)^{64}$$From the Geometric Progression. However, note that as t^{64} is a function that is increasing if t>0, but decreasing if t<0, we have that$$(x+1)^{64}=(x-1)^{64} \iff (x+1)= \pm (x-1)$$So x=0 as x-1 \neq x+1. You have to solve for$$\biggl(\frac{x+1}{x-1}\biggr)^{\!64}=1,\quad x\ne 1$$So set u=\dfrac{x+1}{x-1} and solve \;u^{64}=1:$$u=\mathrm e^{\tfrac{ik\pi}{32}},\enspace\text{whence}\enspace x=\frac{u+1}{u-1}=-i\cot\frac{k\pi}{64}\quad (1\le k\le63).$$The only real solution is x=0, and it corresponds to the case k=32. Note first of all that x=-1 is not a solution. This allows you to write the equation as:$$\left(\frac{x+1}{x-1}\right)^{64}=1.$$There are 64, 64th roots of unity, given by \zeta^k for \zeta a principal root and k=0,1,\dots,63. This gives$$x_k=\frac{1+\zeta^k}{1-\zeta^k}.$$You might be concerned that$x_k$could be real even if$\zeta^k$isn't. It is an exercise to show that this can't happen (assume$(1+z)/(1-z)\in \mathbb{R}$and show that$z\in \mathbb{R}$). Therefore$\zeta^k=\pm 1$. Clearly$\zeta_k=1$doesn't work and so we must have$\zeta^k=-1$and so$x=0$. It gives$x-1=-(x+1)$and$x=0$or$x-1=x+1$, which is impossible. • Surely$x-1 = \pm (x+1)\$, no? – higgs Feb 13 '17 at 13:57
• @Bacon Yes! We have two cases. – Michael Rozenberg Feb 13 '17 at 14:04