Repeated Roots

Let $y = \sqrt{x+ \sqrt{x+ \sqrt{x+\cdots}}}$ . Then we have:

$\begin{aligned} y & = \sqrt {x+y} & \small \color{#3D99F6} \text{Squaring both sides} \\ y^2 & = x+y & \small \color{#3D99F6} \text{Rearranging} \\ y^2 - y - x & = 0 & \small \color{#3D99F6} \text{Solving the quadratic of }y \\ y & = \frac {1 + \sqrt{1+4x}}2 & \small \color{#3D99F6} \text{Note that }y > 0 \end{aligned}$

Note that $y$ is integral when $1+4x$ is a perfect square. Note that $1+4x$ is odd therefore its root $\sqrt{1+4x}$ is also odd, then $\dfrac {1+\sqrt{1+4x}}2$ is integral whenever $1+4x$ is a perfect square.

Next we note that all positive odd perfect squares can be expressed as $4x+1$ . Proof: Let $k$ be any non-negative integer, then any positive odd perfect square is $(2k+1)^2 = 4{\color{#3D99F6}(k^2+k)}+1 = 4{\color{#3D99F6}x}+1$ , where $\color{#3D99F6}x =k^2+k$ .

Then the number of $x$ between 1 and 1000 such that $y$ is integral is given by:

$\begin{aligned} 1 < & \ x < 1000 \\ 1 < k^2 & + k < 1000 \\ 2 \le & \ k \le 32 \end{aligned}$

Since one $k$ gives one $x$ , we have $\boxed{31}$ integral $x$ 's between 1 and 1000 such that $y$ is also integral.

Let $y = \sqrt{x + \sqrt{x + \sqrt{x + \cdots}}} = \sqrt{x + y}$ . Square both sides to find that $y^2 = x + y \ \ \ \ \therefore\ \ \ \ \ x = y(y-1).$ $x$ is an integer whenever $y$ is an integer, and the function $x(y)$ is strictly increasing for $y \geq 2$ . The condition $1 \leq x \leq 1000$ requires $2 \leq y \leq 32$ . (Note that $32\times 31 = 992$ but $33 \times 32 = 1056$ .)

Therefore there are $32 - 2 + 1 = \boxed{31}$ values of $y$ that result in an acceptable value for $x$ .

Let $f(x) = \sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\dots}}}}}$ . Then, if $f(x)$ exists (and it is easy to prove it exists by the monotone convergence theorem) then the relationship $f(x) = \sqrt{x + f(x)}$ must hold. Then $f(x)^2 - f(x) = x \implies (f(x) - \frac{1}{2})^2 = x + \frac{1}{4} \implies f(x) = \sqrt{x + \frac{1}{4}} + \frac{1}{2} = \frac{\sqrt{4x + 1} + 1}{2}$ .

One important thing to note is that I ignore the possibility of a negative radical because from the given definition we know that for positive $x$ , $f(x)$ has to be positive as well. After this, we have a new definition for our function which is much easier to manipulate. It is also much easier to see when $f(n)$ will be an integer. We have a radical in our expression and it is necessary for that radical to evaluate to an integer so we have $\sqrt{4n + 1} = m \implies 4n + 1 = m^2$ . This is now no more than a diophantine equation, and a very simple one too. Notice that what it says is that $m^2 \equiv 1 \mod 4 \implies m \equiv -1,1 \mod 4$ . So $m = 4k - 1$ or $m = 4k+1$ . In the first case we would have $n = \frac{(4k - 1)^2 - 1}{4} = \frac{4k(4k-2)}{4} = 4k^2 - 2k$ . We can easily solve the inequality $1 \leq 4k^2 - 2k \leq 1000 \implies 1 \leq k \leq 16$ so in this case there are $16$ such numbers.

In the other case we would have $n = \frac{(4k + 1)^2 - 1}{4} = 4k^2 + 2k$ and solving the inequality $1 \leq 4k^2 + 2k \leq 1000 \implies 1 \leq k \leq 15$ so there are 15 such numbers. This adds up to $16 + 15 = 31$ such numbers.

There is still an extra constraint though. It must be that $\sqrt{4x + 1} + 1$ is divisible by 2. This is always the case because $4x + 1 \equiv 1 \mod 2$ which implies its square root is also $\equiv 1 \mod 2$ and then if we add $1$ to it we get an even number. So all of our $31$ candidates work, making the solution exactly $31$ .

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

$y=\sqrt{x+y}$

$y^2=x+y$

$x = y^2-y$

Now we have $x$ as a function of $y$ . At this point, we are looking for integer values of $x$ between 1 and 1000. Any integer of $y$ yields an integer for $x$ , and since $y^2-y$ is strictly increasing, we simply need to find the lower and upper limits of $y$ to determine the number of values of $x$ between 1 and 1000.

$y=1$ yields $x=0$

$y=2$ yields $x=2$

So the lower limit of y is 2.

A good starting point to find the upper limit is $\lceil\sqrt{1000}\rceil$ , since $1000-\sqrt{1000} < 1000$ , but close to the upper limit of $y$ .

$\lceil\sqrt{1000}\rceil = 32$

$y=32$ yields $x=992$

$y=33$ yields $x=1056$

Therefore, $2 \le y \le 32$ .

The number of integer values of $x$ in the range $1 \le x \le 1000$ which satisfy the conditions of the problem are $\boxed{31}$ . We do not need to calculate the possible values of $x$ , other than those used to determine the limits.

Well I wrote my solution similar to Sir @Chew-Seong Cheong too however, I tried to reduce it a bit so that it can be done with in mind . So let's make some observations for $x =2,3,4,\cdots$ as below.

For $x=2$ $\begin{aligned} & 2 = \sqrt{2+\sqrt{2+\sqrt{+\sqrt{2+\cdots}}}}\\& 2 =\sqrt{2^2 -2 +\sqrt{2^2-2+\sqrt{2^2-2+\cdots}}} = \sqrt{2×1 + \sqrt{2×1 +\sqrt{2×1+\cdots }}} \\& \end{aligned}$ Similarly for $x=3$ $\begin{aligned} & 3 = \sqrt{6+\sqrt{6+\sqrt{6+\cdots}}}\\& 3=\sqrt{3^2-3+ \sqrt{3^2-3+ \sqrt{3^2 -3+\cdots}}} = \sqrt{3×2 +\sqrt{3×2 +\sqrt{3×2+\cdots}}}\end{aligned}$ So in general for $x$ we can clearly note that $\begin{aligned} & x = \sqrt{x(x-1) +\sqrt{x(x-1) +\sqrt{x(x-1)+\cdots}}}\\&\end{aligned}$ Shows that $x$ is an integer iff any integer nested is the product of consecutive integers where one of the integer is equal $x$ .

So between 1 to 1000 we can have $\begin{aligned} &1< x \leq 1000 \\& \implies 1< x(x-1)\leq 1000\\& \implies 1< 32\times (32-1) \leq 1000 \\& \implies1 < 32\times 31 \leq 1000\\& \implies 2\leq x \leq 32\end{aligned}$ Hence there are exactly $31$ integers for $x$ between 1 too 1000

The integers that satisfy this equation are always right between consecutive squares.

So we can state that y will be an integer if $y = \lfloor(x^2 + (x+1)^2)/2\rfloor$ where x is an integer

In the example: 6 lies between $2^2$ and $3^2$ . Or in other words $6 = \lfloor(2^2 + 3^2)/2\rfloor$

The highest value of $x^2$ so that $x^2$ < 1000 is 31

$n = \sqrt{x+\sqrt {x+\ldots}} \implies n^2 - x = \sqrt{x+\sqrt {x+\ldots}} \implies n^2 - x = n\implies n^2 - n -x = 0$

Polynomial $n^2 - n - x$ in $\mathbb Z[n]$ has integer root if and only if it can be factored over $\mathbb Z$ , i.e., if and only if there exist integers $a$ and $b$ such that $n^2 - n - x = (n-a)(n-b)$ . Expanding the RHS and comparing coefficients gives us $a + b = 1$ and $x = - ab$ which implies $x = a(a-1)$ . It is not loss of generality to assume that $a$ is positive (at least one of $a$ , $b$ is). Thus, the question is how many integers in $[1,1000]$ are product of two consecutive positive integers. Since $31\cdot 32 < 1000 < 32 \cdot 33$ , we have that $x\in \{ 1\cdot 2, 2\cdot 3,\ldots, 31\cdot 32\}$ , so there are $31$ integer values of $x$ .

$3 = \sqrt {6+\sqrt {6+\sqrt {6+\sqrt {6+ \ldots }}}}$

Notice,

$3 = \sqrt {6+3}$ $\implies 3= \sqrt {9}$

So, we have look for square roots which are integers from $1 \rightarrow 1000$ . So, let's start.

$\sqrt {1000} = 10 \sqrt {10}$

$10\sqrt {10}$ is not an integer. By trying multiple times, I finally came to $961$ .

$\sqrt {961} = \boxed {31}$

Now we have our answer. I know that it's not a smart way of solving questions.

Let $\mathbb{N}$ be the set of positive integers.

Let $(x \in \mathbb{N}) \wedge (1 \leq x \leq 1000)$ and let $y = \sqrt{x + \sqrt{x + \sqrt{x + ... }}} \implies y^2 - x = \sqrt{x + \sqrt{x + \sqrt{x + ... }}} = y \implies y^2 - y - x = 0 \implies y = \dfrac{1 \pm \sqrt{4x + 1}}{2}$

$(x \in \mathbb{N}) \wedge (1 \leq x \leq 1000) \implies y = \dfrac{1 - \sqrt{4x + 1}}{2} < 0$ $\therefore$ choose $y = \dfrac{1 + \sqrt{4x + 1}}{2} > 0$ for $(1\leq x \leq 1000) \implies 4x + 1 = j^2 \implies x = \dfrac{j^2 - 1}{4}$ for positive $j$ odd $\implies (1 \leq x = \dfrac{(2k + 1)^2 - 1}{4} = k(k + 1) \leq 1000) \implies k = \lfloor{\dfrac{-1 + \sqrt{4001}}{2}}\rfloor = \boxed{31}$

Y=√(X+Y) Our condition is what both X & Y should be integers and the value of X should be in between 1 to 1000
1=√(0+1) but here our X value is 0 (i.e)<1. So neglect this expression
2=√(2+2)
3=√(6+3) which is given expression in question
4=√(12+4)
5=√(20+5)
6=√(30+6)
7=√(42+7)
8=√(56+8)
9=√(72+9)
10=√(90+10)
11=√(110+11)
12=√(122+12)
13=√(156+13)
14=√(182+14)
See in all the above expressions both the values of X & Y are integers
Think how long it will continue?
30x30=900
31x31=961 where we can get ----> 31=√(930+31)
32x32=1024 where we can get 32=√(992+32)
33x33=1089 where we can get 33=√(1056+33) but here our X value > 1000 so neglect this expression also
So from 1 to 33, neglect 1st & 33rd expressions
so we can get 31 such expression