# 2017 Indonesia MO Problems/Problem 4

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## Problem

Determine all pairs of distinct real numbers $(x, y)$ such that both of the following are true: \begin{align*} x^{100} - y^{100} &= 2^{99} (x-y) \\ x^{200} - y^{200} &= 2^{199} (x-y) \end{align*}

## Solution

Let $x = 2a$ and $y = 2b,$ resulting in the below system. \begin{align*} a^{100} - b^{100} &= a-b \\ a^{200} - b^{200} &= a-b \end{align*} Substitution and factoring of difference of squares results in \begin{align*} a^{100} - b^{100} &= a^{200} - b^{200} \\ a^{100} - b^{100} &= (a^{100} - b^{100})(a^{100} + b^{100}) \\ 0 &= (a^{100} - b^{100})(a^{100} + b^{100} - 1) \end{align*} By the Zero Product Property, either $a^{100} = b^{100}$ or $a^{100} + b^{100} = 1.$ If $a^{100} = b^{100}$ and $a \ne b,$ then $a = -b.$ However, a quick check reveals this to be an extraneous solution.

Thus, $a^{100} + b^{100} = 1.$ If $a = 0,$ then $b = 1,$ and if $b=0,$ then $a=1.$ Now assume $|a|, |b| < 1.$ Because $a^{100} - a = b^{100} - b,$ we have $\frac{a}{b} = \frac{b^{99} - 1}{a^{99} - 1}.$ Since $\frac{b^{99} - 1}{a^{99} - 1}$ is positive, $a$ and $b$ must have the same sign.

Note that $\frac{a^{100} - b^{100}}{a-b} = 1 = \sum_{i=0}^{99} a^i b^{99-i}.$ If $a,b$ are negative, then the sum is negative since each term is negative, so $a,b$ must both be positive. That means $a^{99} + b^{99} < 1,$ so \begin{align*} a^{99} + b^{99} &< a^{100} + b^{100} \\ a^{99} + b^{99} &< a \cdot a^{99} + b \cdot b^{99}. \end{align*} Since $a,b < 1,$ we have $a^{99} + b^{99} < a^{99} + b^{99}.$ However, that can not happen, so there are no more solutions.

This means that the only solutions are $\boxed{(0,2),(2,0)}.$

## See Also

 2017 Indonesia MO (Problems) Preceded byProblem 3 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 Followed byProblem 5 All Indonesia MO Problems and Solutions
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