10.8. Find all pairs of distinct real numbers \(x\) and \(y\) such that \(x^{100} - y^{100} = 2^{99}(x - y)\) and \(x^{200} - y^{200} = 2^{199}(x - y)\).

(I. Bogdanov)
Answer. $(x, y)=(2,0)$ and $(x, y)=(0,2)$.

Solution. For convenience, let's make the substitution $x=2a$ and $y=2b$. Then from the condition we have $(2a)^{100}-(2b)^{100}=2^{99} \cdot (2a-2b)$ and 
$(2a)^{200}-(2b)^{200}=2^{199} \cdot (2a-2b)$. Dividing both equations by powers of two, we get $a^{100}-b^{100}=a^{200}-b^{200}=a-b \neq 0$. Dividing the second expression by the first, we get $a^{100}+b^{100}=1$; hence, each of the numbers $a$ and $b$ in absolute value does not exceed 1.

If $b=0$, then $a^{100}=a$, from which $a=1$. Similarly, if $a=0$, then $b=1$; this leads to two solutions $(x, y)=(2,0)$ and $(x, y)=(0,2)$.

Now let $ab \neq 0$; then $|a|,|b|a^{100}+b^{100}=1$. Contradiction

Remark. After obtaining the inequality (*), the solution can be completed in different ways - for example, using the inequality

$$
\left(a^{99}+a^{98}b+a^{97}b^{2}+\ldots+b^{99}\right)^{100}>\left(a^{100}+b^{100}\right)^{99}
$$

valid for $ab>0$. This inequality can be proven by expanding the brackets and establishing that the coefficient of any monomial on the left is not less than the coefficient of the same monomial on the right.

Comment. Only the answer - 0 points.

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