Lemma: $x,y,\sqrt{x}-\sqrt{y}\in\mathbb Q\in\mathbb Q\implies \sqrt{x},\sqrt{y}\in\mathbb Q$

Since $x_i\in\mathbb Z$ we know that $\forall x_i\ \exists y_i\in\mathbb Z\text{ such that }x_i=y_i^2$ And our recursion becomes $y_i=\sqrt{y_{i-1}-y_{i-2}}<y_{i-1}$ But then $y_{i+1}=\sqrt{y_{i}-y_{i-1}}\not\in\mathbb R$, a contradiction. Edit: wait are you sure the problem is written correctly? It seems that even discarding the rational/integer stuff $x_i=\sqrt{x_{i-1}}-\sqrt{x_{i-2}}<x_{i-1}$, which will lead to a contradiction...

@above I think the problem doesn't say that $x_i\in \mathbb{Z}$, it just $x_i>0$

http://artofproblemsolving.com/community/c6h1217717p6071050 http://artofproblemsolving.com/community/c6h399294p2282914

Assume that there exists a sequence with the problem condition . İt`s obvious that the sequence $x_n$ is increasing. So, $x_{n+1}<\sqrt{x_{n+1}}- \sqrt{x_n}$ for all positive integers $n$. This yields that $x_{n+1}$ is smaller than 1 for all $n$. Summing up the terms $x_3,x_4, \dots,x_{n+2}$ we get that the sum is equal to $\sqrt{x_{n+1}}-\sqrt{x_1}< \sqrt{x_{n+1}}<1$. However, on the other hand, the sum is bigger than $nx_1$. But, as $n$ get closes to infinity, at some point $nx_1$ should be bigger than $1$. So no such sequence exists.