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.
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1$\begingroup$ Does this require $x$ and $y$ to be integers? $\endgroup$– mvwFeb 9, 2015 at 12:09
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1$\begingroup$ Yes,both $x,y$ are integers. $\endgroup$– kingW3Feb 9, 2015 at 12:10
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$\begingroup$ 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. $\endgroup$– jimjimFeb 9, 2015 at 12:10
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2$\begingroup$ I just realized I am the 1000th viewer of this question... $\endgroup$– AvZFeb 10, 2015 at 12:06
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4 Answers
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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 $k=0$.
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$\begingroup$ It's pretty boring when y has to be an integer. $\endgroup$– smciFeb 9, 2015 at 23:38
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$\begingroup$ 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. $\endgroup$– imallettFeb 10, 2015 at 6:34
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$\begingroup$ @imallett: The question says integer solutions (and comments explain this applies to both $x$ and $y$). $\endgroup$ Feb 10, 2015 at 12:52
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$\begingroup$ @MarcvanLeeuwen I knew I was missing something. Thanks! $\endgroup$– imallettFeb 10, 2015 at 16:22
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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.
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$$a_{n+1}-a_n=\frac{x}{\sqrt{a_n+x}+a_n}>0$$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.
answered Feb 9, 2015 at 12:23
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$\begingroup$ Well the problem is the last inequality has $\sqrt{x}$ instead of $x$ since for example $n=2$ and $x=4$ we have that $\sqrt{x+\sqrt{x}}=\sqrt{4+2}=\sqrt{6}<4$ $\endgroup$– kingW3Feb 9, 2015 at 12:32
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$\begingroup$ Oh, yes, I am editing the answer. $\endgroup$ Feb 9, 2015 at 12:35
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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.
