A sequence of nested radicals and its limit

Question

Please Help me in the following Problem

What is the Number Of Natural Numbers ,$n\le30$ for which $\sqrt{n+\sqrt{n+\sqrt{n+\cdots}}}$ is also a prime number.

The only way I am able to find to solve this is calculate each and every term once but it will be extremely lengthy. Please suggest me a shorter way

NOTE This is a NSEJS (STAGE-1) Problem

Answer 1

If $n = p^2 -p$, then $p^2 =n + p $, so $p = \sqrt{n + p}= \sqrt{n + \sqrt{n+p}}$, etc.

Edit: My original answer, above, left a lot to the imagination. The real work would be in analyzing the convergence of the sequence of nested radicals. To this end, fix a positive number $n$, not necessarily an integer for now. Define $a_k = a_k(n)$ by $a_1 = \sqrt{n}$, and $a_{k+1} = \sqrt{n + a_k}$ for $k \ge 1$.

Claim:

1. For fixed $k$, $a_k(n)$ is an increasing function of $n$.
2. For fixed $n$, $(a_k(n))_{k \ge 1}$ is an increasing sequence and bounded above, hence convergent to a limit $p = p(n)$.

The first claim can be proved by induction on $k$. For the second claim, note that $a_2 - a_1 > 0$, and $$ (a_{k+1} - a_k)(a_{k+1} + a_k) = a_k - a_{k-1} $$ for $k >1$. By induction on $k$, $a_{k+1} - a_k > 0$ for all $k$. Now assume for the moment that $n \ge 1$. Then $a_k \ge a_1 = \sqrt{n} \ge 1$. Hence $$ (a_{k+1} - a_k) \le (1/2)( a_k - a_{k-1}) $$ for $k >1$. It follow that the sequence $(a_k(n))_{k \ge 1}$ is convergent. But now if $0 < n < 1$, $a_k(n) \le a_k(1) \le \lim_{k\to \infty} a_k(1)$, so again the sequence $(a_k(n))_{k \ge 1}$ is bounded above, and hence convergent. Let $p = p(n)$ denote $\lim_{k\to \infty} a_k(n)$. Again, $p$ is not necessarily an integer let alone a prime. This completes the proof of the claims.

Now fix $n>0$, and let $p = \lim_{k\to \infty} a_k(n)$. Then we have $$ p = \sqrt{n + p}, $$ as follows from $a_{k+1} = \sqrt{n + a_k}$. Therefore $p^2 - p = n$, and thus $p$ is the unique positive root of the quadratic equation $p^2 - p = n$. Conversely, if we start with a positive number $p > 1$ and put $n = p^2 - p$, then the sequence $a_k(n)$ converges to $p$. Thus $p$ is the limit of the sequence of nested radicals $a_k(n)$ if and only if $p^2 - p = n$.

Now we can solve the original question by searching for primes $p$ such that $p^2 - p \le 30$.

Answer 2

Let the limit be $x$. Then $\sqrt{n+x}=x$. Thus $x^2-x-n=0$... So $x=\frac{1\pm\sqrt{1+4n}}2$. $x$ is positive, so $x=\frac{1+\sqrt{1+4n}}2$.

So let's see: $n=6 \implies x=3$

Say $x$ is a prime, $p$. Then $p^2-p-n=0$, so $n=p(p-1)$...

It looks like infinitely many... (as there are infinitely many primes)