Let $x>0$ and let $x_{m} = \sqrt{x+\sqrt{x+\ldots+\sqrt{x}}}$ (with $m$ roots). We first prove that the sequence $(x_{m})_{m=1}^{\infty}$ is positive and strictly increasing. $x_{2} = \sqrt{x+\sqrt{x}} > \sqrt{x} = x_{1} > 0$ Suppose that $x_{k} > x_{k-1} > 0$ for some $k>1$. Then $x_{k+1} = \sqrt{x+x_{k}} > \sqrt{x+x_{k-1}} = x_{k} > 0$. The claim follows by induction. We now prove that it is bounded above ${x+1 > 2\sqrt{x} > \sqrt{x} = x_[1} > 0$ Suppose that $x+1 > x_{k} > 0$ for some $k \ge 1$. Then ${x+1 = \sqrt{x^2+2x+1} > \sqrt{2x+1} > \sqrt{x+x_{k}} = x_[k+1} > 0$ The claim follows by induction Then the sequence converges to a limit $L>0$ $L = \lim_{k \rightarrow \infty}x_{k} = \lim_{k \rightarrow \infty}\sqrt{x+x_{k-1}} = \sqrt{x+ \lim_{k \rightarrow \infty}x_{k-1}} = \sqrt{x+L}$ Then $L^2 - L -x = 0$ and $L > 0$. Then $L=\frac{1+\sqrt{1+4x}}{2}$ $x_{k} < n$ for every $k \ge 1$ iff $L \le n$ (Since the sequence is strictly increasing and convergent). $L \le n \Leftrightarrow \frac{1+\sqrt{1+4x}}{2} \le n \Leftrightarrow x \le \frac{(2n-1)^2-1}{4} = n^2-n$ Then the positive inteqers $x$ that satisfy are those such that $1 \le x \le n^2-n$