see here: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=56&t=401073&

Goutham wrote: Prove that there exist integers $a, b, c$ all greater than $2011$ such that \[(a+\sqrt{b})^c=\ldots 2010 \cdot 2011\ldots\] [Decimal point separates an integer ending in $2010$ and a decimal part beginning with $2011$.] Gist of my solution: We have that the sequence $( P(x) )_{x \ge 1}$ is equidistributed modulo 1, if P(x) is a polynomial having leading coefficient irrational, by Van der Corput and Kronecker's Theorem. Now, apply this to $P(x) = tx^c$, where $t = \frac 1 {10000} (m+\sqrt n)^c$, for some $m,n,c \ge 2011$ to get that there exists $x \in \mathbb{N} $ such that ${P(x) } = 0.20102011 \cdots $. But then, we have that $a=mx, b = nx^2 , c=c$ suffice.

Rijul saini wrote: Goutham wrote: Prove that there exist integers $a, b, c$ all greater than $2011$ such that \[(a+\sqrt{b})^c=\ldots 2010 \cdot 2011\ldots\][Decimal point separates an integer ending in $2010$ and a decimal part beginning with $2011$.] Gist of my solution: We have that the sequence $( P(x) )_{x \ge 1}$ is equidistributed modulo 1, if P(x) is a polynomial having leading coefficient irrational, by Van der Corput and Kronecker's Theorem. Now, apply this to $P(x) = tx^c$, where $t = \frac 1 {10000} (m+\sqrt n)^c$, for some $m,n,c \ge 2011$ to get that there exists $x \in \mathbb{N} $ such that ${P(x) } = 0.20102011 \cdots $. But then, we have that $a=mx, b = nx^2 , c=c$ suffice. can someone prove it in a Olympiad's way? simplfied into $\sqrt{k}x^n$ $(n \in N)$ is enough for me.