Bogdanov I.i.

Find all such pairs of distinct real numbers $x$ and $y$ that $x^{100}-y^{100}=2^{99}(x-y)$ and $x^{200}-y^{200}=$ $2^{199}(x-y)$
For convenience, let's make the substitution $x=2a$ and $y=2b$. Then $a^{100}-b^{100}=a^{200}-b^{200}=a-b \neq 0$. 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 $b=1$.

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

## Answer

$(2,0)$ and $(0,2)$.