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

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? ............

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    this is some moscow olympiad or something – 2017-02-13
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    @dp1611 got it right. – 2017-02-13

4 Answers 4

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It gives $x-1=-(x+1)$ and $x=0$ or $x-1=x+1$, which is impossible.

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    Surely $x-1 = \pm (x+1)$, no? – 2017-02-13
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    @Bacon Yes! We have two cases. – 2017-02-13
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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 $k=32$.

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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 $x=0$.

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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 $x-1 \neq x+1$.