25
$\begingroup$

Find integer solutions to the equation $$\underbrace{\sqrt{x+\sqrt{x+\sqrt{x+\cdots+\sqrt{x+\sqrt{x}}}}}}_{1000}=y$$ The first thing that popped to my mind was the infinitely nested radical and the recurrence relation $$a_{n+1}=\sqrt{x+a_n}$$ That lead me nowhere so I tried to show there aren't any such $(x,y)$ except for the trivial $(0,0)$ I think I proved that $$\sqrt{a_0}\leq a_{n}\leq \sqrt{a_0}+1$$ but the proof itself was pretty vague and even if it is correct doesn't help me.I have no idea what else to do,any hints would be appreciated.

$\endgroup$
  • 1
    $\begingroup$ Does this require $x$ and $y$ to be integers? $\endgroup$ – mvw Feb 9 '15 at 12:09
  • 1
    $\begingroup$ Yes,both $x,y$ are integers. $\endgroup$ – kingW3 Feb 9 '15 at 12:10
  • $\begingroup$ This would be close enough to the question with indefinitely many iterations, the possibility of a explicit answer is slim, should settle for the approximate one. $\endgroup$ – Arjang Feb 9 '15 at 12:10
  • 2
    $\begingroup$ I just realized I am the 1000th viewer of this question... $\endgroup$ – AvZ Feb 10 '15 at 12:06
50
$\begingroup$

If any of the $a_n$ fails to be an integer, then $a_{n+1}$ will also fail. So especailly $x$ must be a square, say $x=k^2$; and also $x+k= k(k+1)$ must be a square. Since $\gcd(k,k+1)=1$, this means that $k$ and $k+1$ must both be square. This allows only $k=0$.

$\endgroup$
  • 3
    $\begingroup$ This is very neat +1. $\endgroup$ – kingW3 Feb 9 '15 at 12:33
  • $\begingroup$ It's pretty boring when y has to be an integer. $\endgroup$ – smci Feb 9 '15 at 23:38
  • $\begingroup$ Let $a_0=\sqrt{x}=\frac{1}{2}(9-\sqrt{17})$, then $a_1=\sqrt{x+a_0}=\sqrt{x+\sqrt{x}}=2$. So $a_0$ is not an integer, but $a_1=2$ very much is. Calculation. $\endgroup$ – imallett Feb 10 '15 at 6:34
  • $\begingroup$ @imallett: The question says integer solutions (and comments explain this applies to both $x$ and $y$). $\endgroup$ – Marc van Leeuwen Feb 10 '15 at 12:52
  • $\begingroup$ @MarcvanLeeuwen I knew I was missing something. Thanks! $\endgroup$ – imallett Feb 10 '15 at 16:22
3
$\begingroup$

Another problem where starting with the simpler version of the problem does wonders:

$\sqrt{x+\sqrt{x}}=y$

We need both $x+\sqrt{x}$ and $\sqrt{x}$ to be squares; say $x=k^2$, $k\in \mathbb{N}$.

Then $x+\sqrt{x}=k^2+k<(k+1)^2$, so it cannot be a perfect square.

Taking further square roots is not going to remedy this.

$\endgroup$
1
$\begingroup$

$$a_{n+1}-a_n=\frac{x}{\sqrt{a_n+x}+a_n}>0$$So, $\{a_n\}$ is an increasing sequence. Now, we know that $a_n\to \frac{1+\sqrt{1+4x}}{2}$ Observe that (You've already observed it :-))$$\sqrt{x}+1-\frac{1+\sqrt{1+4x}}{2}=\frac{2\sqrt{x}+1-\sqrt{1+4x}}{2}=\frac{4\sqrt{x}}{(2\sqrt{x}+1+\sqrt{1+4x})}>0$$Thus $\forall n\ge 1$ $$\sqrt{x}<a_n\le \frac{1+\sqrt{1+4x}}{2} <\sqrt{x}+1$$ where $a_0=\sqrt{x}$ So if $x$ is a square integer $a_n$ cannot be an integer. If $x$ is not square, then $a_0=\sqrt{x}$ is irrational and hence all $a_n,\ n\ge 1$ are irrational. So in any case, $a_n$ cannot be an integer.

$\endgroup$
  • $\begingroup$ Well the problem is the last inequality has $\sqrt{x}$ instead of $x$ since for example $n=2$ and $x=4$ we have that $\sqrt{x+\sqrt{x}}=\sqrt{4+2}=\sqrt{6}<4$ $\endgroup$ – kingW3 Feb 9 '15 at 12:32
  • $\begingroup$ Oh, yes, I am editing the answer. $\endgroup$ – Samrat Mukhopadhyay Feb 9 '15 at 12:35
1
$\begingroup$

If we assume $x, y \in \mathbb{N}_0$ and write the task like this \begin{align} F_1 &= x \\ (*) \quad F_n &= x + \sqrt{F_{n-1}} \quad \left( n \in \{ 2, \ldots, 1000 \} \right) \\ F_{1000} &= y^2 \end{align} we see that $F_{1000}$ must be a non-negative integer, because $y$ is an integer. But also $\sqrt{F_{999}}$, because $$ F_{1000} = x + \sqrt{F_{999}} $$ and $F_{1000}$ and $x$ are integers. If $\sqrt{F_{999}}$ is an integer then $F_{999} = k_{999}^2$ for some non-negative integer $k_{999}$.

By the same argument $$ F_{n} \in \mathbb{N}_0 \Rightarrow \sqrt{F_{n-1}} \in \mathbb{N}_0 \Rightarrow F_{n-1} = k_{n-1}^2 \in \mathbb{N}_0 $$ for $n \in \{ 3, \ldots, 1000 \}$. It also follows from the first above equations that $\sqrt{F_1} = k_1 \in \mathbb{N}_0$.

Between the squares $k_i^2$ equation $(*)$ gives the relation $$ k_i^2 = k_1^2 + k_{i-1} = 2 k_1^2 + k_{i-2} = (i-1) k_1^2 + k_1 $$ which gives troubles already for $$ k_2^2 = k_1^2 + k_1 = k_1(k_1 + 1) $$ because either $k_2 = k_1 = 0$ or otherwise $k_2$ would have to be the geometric mean of the integers $k_1$ and $k_1 + 1$ which is not an integer: $$ k_1 = \sqrt{k_1 k_1} < \sqrt{k_1(k_1+1)} < \sqrt{(k_1+1)(k_1+1)} = k_1 + 1 $$ This leaves only the trivial solution $y = x = 0$.

So the nice $$ y = 999 k_1^2 + k_1 $$ gives nothing interesting.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.