Problem 609

Solution:

We will show that such numbers $a, b$ and $c$ exist. The number $x=(a+\sqrt{b})^{c}+(a-\sqrt{b})^{c}$ is an integer. It is sufficient to choose $a, b, c$ so that $x$ is divisible by $10^{4}$ and $7989,7989> (a-\sqrt{b})^{c}>7989,7988$.

For odd $c$, the number $x=2 a^{c}+2\binom{c}{2} a^{c-2}+\cdots+2\binom{c}{c-1} a$ is divisible by $a$, so it suffices to take $a$ which is divisible by $10^{4}$. We achieve the second condition by choosing $a$ and $b$ so that $1<a-\sqrt{b}<\sqrt{\frac{7989,7989}{7989,7988}}$ — for example, $a=10^{8}$ and $b=(a-1)^{2}-1$. Indeed, let $c$ be the smallest odd natural number for which $(a-\sqrt{b})^{c}>7989,7988$. Such $c$ is obviously greater than 2011 (in this case $c=1797184159)$ and $(a-\sqrt{b})^{c}<7989,7989$.