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Find the number of positive integers $n$ such that $\sqrt{n+\sqrt{n+\sqrt{n}}}<10$ for any finite number of square root signs.

I know something with squaring and repeating, but what does "finite number of square root signs" mean?

InanimateBeing
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D S
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    What have you tried? – Parcly Taxel Oct 03 '22 at 13:39
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    On re-reading, I think my first interpretation (now deleted) was incorrect. I now believe it means: find all natural numbers $n$ such that each of $n, \sqrt n, \sqrt {n+\sqrt n}, \sqrt {n+\sqrt {n+\sqrt n}}, \cdots$ is less than $10$. The sequence, to be clear, is $a_i$ where $a_1=n$ and $a_{i+1}=\sqrt {n+a_i}$ – lulu Oct 03 '22 at 13:41
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    @lulu I don't think that the $0$-root expression would be $n$. Logically continuing it based on the recurrence would give the zeroth term as $0$. – Parcly Taxel Oct 03 '22 at 13:44
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    @ParclyTaxel Yes, I agree. My sequence should begin with $a_0=0$, though the recursion, $a_{i+1}=\sqrt {n+a_i}$ was correct. This makes $a_1=\sqrt n$ and $n$ itself does not appear in the sequence at all. – lulu Oct 03 '22 at 13:48
  • @DS I have rollback-ed your edit as 1. Hopefully your doubt has been resolved. 2. Such edits aren't appreciated. 3. Correct way is to ask the users like you already did and then wait a bit (some times it's even more than a day) to get the reply. 4. If you edit question in such a way that now the answer would require a different answer to be posted since the question is changed that much then the correct thing to do instead is to post a new question. For example: see this post. – InanimateBeing Oct 03 '22 at 16:14

2 Answers2

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It seems that the question means to ask you, as mentioned by the user lulu, that:
Find all natural numbers $n$ such that each of $\sqrt n,\sqrt{n+\sqrt n}, \sqrt{n+\sqrt{n+\sqrt n}},\dots$ is less than 10.
Which is a round-about way of asking:
Find all natural numbers $n$ such that $\sqrt{n+\sqrt{n+\sqrt {n+\dots}}}<10$.

To solve it, we may use the standard operating procedure:
$$\text{Let }\sqrt{n+\sqrt{n+\sqrt {n+\dots}}}=x$$ $$\Rightarrow\sqrt{n+x}=x$$ $$\Rightarrow x^2-x-n=0$$ $$\Rightarrow x=\frac{1\pm\sqrt{1+4n}}{2}$$ And we are given that $x<10$. So: $$\frac{1\pm\sqrt{1+4n}}{2}<10$$ $$\pm\sqrt{1+4n}<19$$ $$1+4n<361$$ $$n<90$$

But for an infinite sequence would $n=90$ not be less than $10$. Thus, for a finite sequence $n=90$ shouldn't be a problem.
Had we got something like $n<89.92$ then we would have concluded that max value $n$ can assume is $89$.

Thus, for your case, $n$ can be any value from $1$ to $90$.

InanimateBeing
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  • that matches with the answer scheme, but why should we consider only 90, maybe 91 would work too... – D S Oct 03 '22 at 14:47
  • @DS no it wont. First lets check this way: $\sqrt {91+\sqrt {91}}=10.027$ which does not satisfies the condition. Second, see we are getting a limit of 90 at infinity so we can say 90 will work for finite. But 91 is waaayyy far off from 90 and we can't make that jump. Even if $n$ was real, still we'd get $n\in[0,90]$ for finite root sign as even 90.00000000001 would not satisfy the condition, say for, 1,000,000,000,000 number of square roots. Still considering $n$ to be real, if we had got 6.625 as answer then too only 6.625 would satisfying condition even for a VERY large number (not 6.626). – InanimateBeing Oct 03 '22 at 15:57
  • For $n=90.001$ it is enough to have 5 roots to obtain result more than 10. – Ivan Kaznacheyeu Oct 04 '22 at 07:20
  • thanks @InanimateBeing – D S Oct 05 '22 at 10:36
  • @DS glad I was helpful :) – InanimateBeing Oct 05 '22 at 14:08
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As the number of square roots goes from $1$ to $\infty$, the nested radical will increase monotonically from $\sqrt n$ to a limit of $\frac{1+\sqrt{1+4n}}2$. This means

  • if a choice of $n$ produces a limit greater than $10$, it is inadmissible since some finite number of square roots will also produce a result greater than $10$
  • if a choice of $n$ produces a limit less than $10$, it is admissible; all finite nested radicals of its form will produce lesser results

Now $$\frac{1+\sqrt{1+4n}}2<10\implies n<90$$ This is the tighter bound on $n$, since $\sqrt n<10\implies n<100$. $n=90$ is also admissible though since we deal with a finite number of square roots, not the limit. There are thus $90$ admissible integers.

Parcly Taxel
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