Consider an integer ${n}\ge{2}$. Find all integers $x$ such that, for any number of radicals: $$\sqrt{x + \sqrt{x+ \ldots+\sqrt{x}}}<n$$
I know that this problem is supposed to be solved using induction but I just can't figure out how. Assuming an infinite number of nested radicals I concluded that: $$A=\sqrt{x + \sqrt{x+ \ldots+\sqrt{x}}}\implies A^2 = x + \sqrt{x + \sqrt{x+ \ldots+\sqrt{x}}}\implies A^2 = x+ A$$
Solving for A, we find that the inequality holds for all positive integer values of $x$, such that $x\le{n^2 - n}$, which seems to be in line with the solution.
Edit: I realise that I should have mentioned this earlier, but I was hoping for a clearer answer. This is the answer from my problems book, which I don't really understand:
It is obvious that for $\sqrt{x + \sqrt{x+ \ldots+\sqrt{x}}}$ to exist, we must have $x\ge{0}$, and for it to be smaller than $n$ for any number of radicals it must be $n^2 > x$. By setting $u = n^2 -x$, we notice that $u$ is a positive integer. Consequently, $n^2 < u(u+1)$ and because both of them are positive integers, it follows that $u\ge{n}$ and thus $x\le{n^2 -n}$. Using induction on the number of radicals we conclude that for every positive integer the statement $x\le{n^2 -n}$ true. Therefore $x\in{\{0, 1, \dots, n^2 -n\}}$.
Can you please help me clarify the proof process. The parts I don't understand specifically are how the author arrives at the inequality $n^2 < u(u+1)$ and how he uses induction on the number of radicals.