a=2, b=2011 works by induction, but I'd really appreciate it if someone could post a proof that is not "guess first and prove later". There should be a non-determinist proof, something that proves the existence of a,b without having to determine them exactly.

Concyclicboy wrote: Determine if there are positive integers $a, b$ such that all terms of the sequence defined by $ X_{1}= 2010 $, $ X_{2}= 2011 $, $ X_{n+2}= X_{n}+ X_{n+1}+a\sqrt{X_{n}X_{n+1}+b}$, $ n\geq 1 $, are integers. writing $x_nx_{n+1}=u_n^2-b$, with $u_n\ge 0$, we get : $x_{n+2}=x_n+x_{n+1}+au_n$ $x_{n+1}x_{n+2}=x_nx_{n+1}+x_{n+1}^2+au_nx_{n+1}$ $u_{n+1}^2=x_{n+1}^2+au_nx_{n+1}+u_n^2$ And so obviously, any $(2,b)$ such that $x_1x_2+b$ is a perfect square is a solution. So, any $(a,b)=(2,t^2-2010\cdot 2011)$ with $t\ge 2011$ is a solution. (If we want all solutions, it remains to check if $a=2$ is the only possibility).

I got a different solution. As in the other solutions, I will show that $a=2$ and $b=2011$ works. It's easy to check that $x_3$ will be an integer. Now, we have that \[x_{n+2}-x_{n+1}-x_n = 2\sqrt{x_{n+1}x_n + 2011}\] Squaring this relation, we get that \[x_{n+2}^2+x_{n+1}^2+x_n^2 -2x_{n+2}x_{n+1} - 2x_{n+2}x_n = 2x_{n+1}x_n + 8044\] In this last relation, changing $n$ for $n+1$ and subtracting, we get: \[(x_{n+3}-x_n)(x_{n+3}+x_n) -2x_{n+2}(x_{n+3}-x_n) - 2x_{n+1}(x_{n+3}-x_n) = 0\] As the sequence is strictly increasing, we can cancel the factor $x_{n+3}-x_n$ and obtain that \[x_{n+3}-2x_{n+2}-2x_{n+1}-x_{n} = 0\] This new relation shows that the sequence satisfies a linear recurrence of third order with integer coefficients and as the three first terms of the sequence are integers, all terms must be integers.

The question doesn't ask us to find all (a,b) ??

We will show that if $x_1=k-1, x_2=k$ and $x_{n+2}=x_n+x_{n+1}+2\sqrt{x_nx_{n+1}+k}$ for $n\ge1$ and $k>1$ then every term of the sequence is an integer. Taking $k=2011$ gives the desired result. Consider $y_n=\sqrt{x_nx_{n+1}+k}$. We will show $y_n=x_{n}+y_{n-1}$ for $n>1$. Since $y_1=x_2=k$ are integers, it will follow $y_n$ is always an integer so $x_{n+2}=x_{n}+x_{n+1}+2y_n$ is always an integer. We proceed by induction. Since $y_2=2k=k+k=x_2+y_1$ we have our base case. For our inductive hypothesis suppose $y_{m-1}=x_{m-1}+y_{m-2}$. Now, \[ y_m=\sqrt{x_mx_{m+1}+k}=\sqrt{x_m(x_{m-1}+x_m+2y_{m-1})+k} = \sqrt{(x_m+y_{m-1})^2+k+x_mx_{m-1}-y_{m-1}^2}. \]However, since $y_{m-1}^2=x_{m-1}x_m+k$, this is equivalent to \[ y_m = x_m + y_{m-1}. \]This finishes our induction and thus, the problem. $\blacksquare$

I have two solutions one use induction and fibonacci numbers:

The second is without the "importance" of the first elements on the sequence (and my favorite):

AHZOLFAGHARI wrote: The question doesn't ask us to find all (a,b) ?? No. note that problem only asks "whether there exist such $a,b$". It does not ask to find all $a,b$. In any case, it is an interesting question to find all such $(a,b)$, which I am interested to see a proof of (I can't find one).