An Infinitely Nested Radical

We let $x = \sqrt{n+\sqrt{n+\sqrt{n+...}}}$ , and it is given in the problem that $x$ is an integer. We square both sides of this equation to get $x^2 = n+\sqrt{n+\sqrt{n+\sqrt{n+...}}}$ . The RHS of the squared equation is clearly equal to $n + x$ , and so $x^2 - x = n \to x(x-1) = n$ . The largest integer $n < 1000$ which can be expressed as the product of two consecutive integers is $992 = 32\cdot31$ , and so our answer is $\fbox{992}$ .

We can write the infinitely nested radical as X. Then,

$X = \sqrt{n+\sqrt{n+\sqrt{n+...}}}$ . Squaring both sides will give us:

$X^{2} = n + \sqrt{n+\sqrt{n+\sqrt{n+...}}}$ . Therefore,

$X^{2} = n + X$ , Then, $X^{2} - X = n$ or $X (X - 1) = n$ .

The greatest value for n less than 999 is $32 \times (32 - 1) = 992$

Let this radical be $x$ . If $n$ is added to $x$ and square rooted, we will have $x$ . So $x^2=n+x$ . So by solving for positive $x$ , we get $x=\frac{1+\sqrt{4n+1}}{2}$ . To let this be a positive integer, $4n+1$ must be a odd perfect square. $4(999)+1=3997$ , and the closest odd perfect square smaller than 3997 is 3969, which gives us $n=992$ .

Let $x=\sqrt{n+\sqrt{n+\cdots}}$ . Then we have $x^2=n+x$ , or equivalently $x=\frac{\sqrt{4n+1}+1}{2}$ (because $x>0$ ). Since $x$ is a integer, we must have $4n+1$ is a perfect square. Since $n \le 999$ , we have $4n+1 \le 3997$ . Therefore, we obtain $4n+1=3969$ and $n=992$ . Substitute, $n=992$ , we have $x=32$ is a positive integer. Therefore, $n=992$ .

Let's first set this expression equal to $x$ : $x=\sqrt{n+\sqrt{n+\sqrt{n+ \cdots}}}$ Notice that $\sqrt{n+\sqrt{n+ \cdots}}$ is nested inside the first square root. Therefore, we can substitute $\sqrt{n+\sqrt{n+ \cdots}}$ for $x$ . This yields the equation: $x=\sqrt{n+x}$ . Squaring both sides gives: $x^2=n+x$ . Rearrange: $x^2-x=n$ . Factor: $x(x-1)=n$ . Because we know the original expression is equal to a positive integer, $x$ is a positive integer. Now we want to find a certain $x$ such that $n$ is maximized and $n \leq 999$ . Notice that $31^2 < 1000$ , and $32^2 > 1000$ . Therefore, we want to try $x=32$ to get $31\times 32$ , a number somewhere between $31^2$ and $32^2$ . Sure enough, $31\times 32 = 992$ , just under $1000$ . Therefore, our answer is $\boxed{992}$

Let the value of the expression $\sqrt{ n + \sqrt{n + \sqrt{n + \cdots}}}$ be $\alpha$
then $\alpha^{2}$ = $n + \sqrt{ n + \sqrt{n + \sqrt{n + \cdots}}}$
So,
$\alpha^{2} - \alpha$ = $n$
$\alpha \times (\alpha - 1)$ = $n$
Therefore $n$ is a product of two consecutive numbers.
The largest number $n \leq 999$ is $992$ .
992 = $32 \times (32 - 1)$
992 = $32 \times 31$

Assume that the given nested radical is x, i.e.

x=(n+(n+(n+(n+(n+….)1/2)1/2)1/2)1/2)1/2

This implies that x2=n+x.

x2-x-n=0

The roots of this equation are x = (1+(1+4n)1/2)/2 (Sridaracharya’s formula, considering only the positive
value of the square root.)

For x to be an integer, 4n+1 must be a perfect square, and its square root must be an odd number. Since n<1000, the value of 4n+1 is lesser than 4001. The greatest perfect square lesser than 4001 is 632 = 3969, which satisfies our requirements.

4n+1=3969 implies that n=992.

Therefore, the maximum value of n = 992.

By assuming $$x=\sqrt{n+\sqrt{n+\sqrt{n+\ldots}}}$$,we obtain $$x=\sqrt{n+x} \Rightarrow x^2-x=n\leq999 \Rightarrow x\leq\frac{1+\sqrt{3997}}{2}=32,\ldots$$.We obtain the maximum value of $$x=32 \Rightarrow n=32^2-32=992$$

Let $S = \sqrt{n + \sqrt{n + \sqrt{n + \cdots}}}$ . We want to find another way to write $S$ in terms $n$ to solve for $S$ and $n$ . We can square $S$ and subtract $n$ such that $S^2 = n + \sqrt{n + \sqrt{n + \cdots}},$ $S^2 - n = \sqrt{n + \sqrt{n + \sqrt{n + \cdots}}} = S.$ Hence, $S^2 - n = S$ . We notice that this is a quadratic in $S$ , so we use the quadratic equation $S = \frac{1 \pm \sqrt{1 + 4n}}{2}.$ Since $S$ is the largest positive integer, satisfying these conditions, the discriminant must be the largest odd square in the form $1 + 4n$ where $n \leq 999$ . This is $1 + 4 * 992 = 3969 = 63^2$ . Hence, $S = \frac{1 + \sqrt{1 + 4 * 992}}{2} = 32$ and $n = 992$ .

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

Therefore $x = \sqrt{ n + x }$

i.e $x^2 - x = n$

Now, we have to find $n$ such that $n \le 999$ .

Also, from this equation we could say that $x^2$ should be slightly greater than $999$ .

and since, $\sqrt{999} \approx 31.61$

Thus, we'll need a number greater than $31$ i.e $32$

Therefore , putting $x = 32$ we could find greatest n .

$\Rightarrow 32^2 - 32 = n \Rightarrow 992 = n$

Hence the answer is $992$ . Cheers !

Let $x = \sqrt{n+\sqrt{n+\sqrt{n+...}}}$ . $x$ can also be found inside the radical, therefore $x$ can be expressed as:

$x = \sqrt{n+x}$

$x^2 = n+x$

$x^2 - x - n= 0$

$x = \frac{1+\sqrt{(-1)^2 - 4(1)(-n)}}{2}$

$x = \frac{1+\sqrt{1 + 4n}}{2}$

Since x is a positive integer, $\sqrt{1 + 4n}$ must be an positive odd integer. The highest integer value of n to fulfill this condition is $n = 992$ .

\sqrt{n\sqrt{n\sqrt{n\sqrt{...}}}}=x quadratic LHS and RHS x^2-x-n=0, we will get n=31x32=992

Let S be equal to the infinitely nested radical expression. Squaring S we see that: S^2 = n + S This is because when you square an infinitely nested expression that continually diminishes, the last term is insignificant and the left over radical will also be an infinitely nested expression equal to the initial S. Then, moving everything to one side we get: S^2 - S - n = 0 Let a and b be the two roots to this quadratic. Then we can write it as: (S-a)(S-b) = 0 because n is positive, we know that one of a or b has to be negative and WLOG we can assume that b is negative and we can replace b with -p where p, along with a, is a positive integer. So, (S-a)(S+p) = 0 so p - a = 1 and p * a = n because we need an n less than 1000 and p and a can only differ by one, we want the biggest consecutive numbers such that their product is less than 1000, and we get p = 32 and a = 31 which gives us n = 992.

Let the nested radical be x. We have x = sqrt (n+x) => x^2-x-n. We have to find largest n < 999 where the above equation has integer roots. This occurs when the discriminant 4n+1 is a perfect square and the largest such n < 999 is n=992

y=sqrt{n+sqrt{n+sqrt{n}}} y^2=n+y y(y-1)=n=an integer 32*31=992 Therefore n=992

Let, y is an integer & by question y= /sqrt {n+{n+{n+....}}}. /Right arrow y^2-y= n. so, for highest possible value for n /leq 999, y^2-y must be also the highest possible value /leq 999. that is 32^2-32=992.

Let the original expression be equal to $x$ , where $x$ is the largest positive integer yielded by the infinitely nested radical, given that $n\leq999$ .

First, note that $x=\sqrt{n+\sqrt{n+\sqrt{n+\ldots}}}$ is equivalent to the expression $x=\sqrt{n+x}$ , due to the infinite nature of the nested radical.

Rearranging by squaring both sides and collecting all $x$ terms gives: $n=x^{2}-x$ .

This equation infers that a value of $n$ must be chosen such that $x$ is the greatest possible integer, and $n\leq999$ .

By inspection, when $x=31, n=32^{2}-32=1024-32=992$ . This is the greatest possible value of $n$ , as $x=33$ gives $x=1056$ , which does not satisfy $x\leq999$ .

Therefore, the answer to the given problem is $n=992$ .

$Q.E.D.$

sqrt{n+sqrt{n+sqrt{n+.......}}} = m, m is a positive integer.

=> sqrt{n+m} = m

Squaring both sides we get, n + m = m^{2} => n = m^{2} - m = m(m - 1)

Since 31^{2} < 999 < 32^{2}

we try out 31 * 32. We get 992 < 999. Obviously 32 * 33 would be bigger than 999 and 30 * 31 is lower than 31 * 32. Thus 992 is the answer.

let a = \sqrt{n+.......} now squaring both the sides we get a^2 = n + \sqrt{n+.......} a^2 = n + a

a^2 -a -n = 0 ................(1)

now we know that n should be maximum and less than or equal to 999 also n is a positive integer

now in quadratic equation (1)[ using middle term split theorem]

-n = -p\timesq and, -1 = -p+q or 1=p-q

now it is clear that n is multiplication of two cosecutive numbers and also so the 32\times31=992 which is maximum and less than 999......

for any other two cosecutive numbers you would get more than 999 or less than 992...

Let x=\sqrt{n+\sqrt{n+\sqrt{n+.... \sqrt{n+x}=x n+x=x^2 x^2-x-n=0 x=\frac {1+\sqrt{1+4n}}{2} It is evident n=y(y+1) \leq 999 After solving for y, the answer is y=31 and therefore n=y(y+1)=31*32=992

notice n = (1+(4x+1)^(1/2))/2 satisfy n^2 = n+x, let the infinte radical equal to x the given condition is similar as to find an integer x such that x = x^2 - n in order maintain the radical an integer every process breaking the radical. but it leads to x = (1+(4n+1)^(1/2))/2 as x is integer , then 4n+1=(2k+1)^2 , k is an integer, and n=992 is the largest integer satisfy the condition for n<=999

Let $S=\sqrt{n+\sqrt{n+\sqrt{n...}}}$ .

It can be seen that $S=\sqrt{n+S}$ , so that $n=S^2-S=S(S-1)$ . This problem then boils down to finding the greatest integer $S$ such that $S(S-1)<1000$ . (S must be an integer as it is stated in the problem.)

Note that $32^2=1024$ is just a little over 1000, so we try 32 as $S$ and find $S(S-1)=32(31)=\boxed{992}$ .

How can we be sure this is the maximum value? $S=33$ does not work as $32^2$ is already greater than $1000$ .

Let $\mathrm{x}$ = $\sqrt{n + \sqrt{n + \sqrt{n + \dots}}}$

$\Rightarrow$ $\mathrm{x}$ = $\sqrt{n + x}$

On squaring both sides,

$x^2$ = $\mathrm{n + x}$

$\Rightarrow$ $x^2$ $\mathrm{-x}$ $\mathrm{-n}$ = $\mathrm{0}$

$\Rightarrow$ x = $\frac{1 \pm \sqrt{4n + 1}}{2}$

For $\mathrm{n}$ = $\mathrm{992}$

$\sqrt{4n + 1}$ = 63

Hence, $\mathrm{x}$ = $\mathrm{32}$ , which is a positive integer

Thus, our answer is $\fbox{992}$

let us take n=999 , then if the value of the integer is 'a', a=sqrt(999+sqrt(999+sqrt(999+............) a^2=999+a a^2-a-999=0 the solution does not give integer , it gives integer value when the constant term can be written as product of two consecutive integers and that is shorter than 1000, here the product of 31*32=992 become the integer.

Notice that each inner radical of $x = \sqrt{n + \sqrt{n + \cdots}}$ is equal to $x$ itself.

Thus $x = \sqrt{n + x}$ , thus $x^2 - x = n$ .

The largest integer solution under 1000 is $32^2 - 32 = 992$ .

Let $L = \sqrt{n+\sqrt{n+\cdots}}$ . The hard part is showing convergence, which can be done with induction and the convergence of bounded monotone sequences Manipulating, we get $L^2 - n =L$ , and since $L>0$ , the quadratic formula yields $L = \frac{1}{2}(1+\sqrt{1+4n})$ . Suppose this equals an integer $m$ . Then with a little algebra, we have $n = m^2 -m$ . Since $31<\sqrt{1000}<32$ , we try $m=32,33$ and get $n = 992,1056$ . Thus, 992 is the answer.

From the nested radical

$\sqrt{n+\sqrt{n+\sqrt{n+\cdots}}} = x$

with

$n, x \in \mathbb{N}$

we can deduce that

$\Rightarrow \sqrt{n+x} = x$

Solving for x:

$\Rightarrow x^2 - n - x = 0$

$\Rightarrow x = \frac{1}{2} + \frac{\sqrt{1 + 4n}}{2}$

If we want $x$ to be an integer, the square root in the equation needs to be uneven:

$\Rightarrow \sqrt{1+4n} = 2k+1, k \in \mathbb{N}$

Solving for n:

$\Rightarrow n = k^2 + k$

Trying some different values for k:

$30^2 + 30 = 930,$

\fbox{31^2 + 31 = 992},

$32^2 + 32 = 1056$

Which shows that $n = 992$

Let a = the expression in question (which I am having trouble formatting correctly so just pretend it's written here. While you're at it, pretend that all the 'a's and 'n's look the same)

Since the part under the second radical is the same as the original expression, the expression can be written as $\sqrt{n + a}$

so a = $\sqrt{n + a}$ .

Squaring both sides produces $a^{2}$ = n + a

Solving for n give us

n = $a^{2}$ - a = a( a-1)

This tells us that a will be an integer if and only if n is the product of two consecutive integers. The largest value of n that fits the conditions is $31 \times 32$ , or 992.

If we take the value of the entire root thing as x, we may write sqrt(n+x) = x or , n+x = x^2.

We have a quadratic equation in x which is a positive integer, hence its discriminant should be a perfect square.

Also as discriminant is 1+4n, which is odd, we may write 1+4n = (2p+1)^2 where p = an integer.

Expanding we have n = p(p+1). So we proved that n is a product of two consecutive positive integers. The next work is trivial: as n is less than or equal to 999, we find that the closest possible value of n is 992, if p = 31. So, answer is 992.

Let the infinite radical chain be equal to $S$ . Since the chain is infinite, we can write $\sqrt{n + S} = S$ . Expanding this yields the quadratic $S^2 - S - n = 0$ . By the quadratic formula, we have an expression for $S$ , $S = \frac{1 \pm \sqrt{1 + 4n}}{2}$ .

Thus we seek a value of $n$ such that $1 + 4n$ is a square of some odd number. Since $63^2 = 3969$ and $65^2 = 4225$ , and $n \leq 999$ , we can solve $1 + 4n = 3969$ to get $n = 992$ . Note that this is largest since $n \leq 999$ and thus $1 + 4n \leq 3997$ .

Hence our answer is $\boxed{992}$ .

Letting the given expression equal $x$ , we obtain

$x=\sqrt{n+\sqrt{n+\sqrt{n+\ldots}}}$ ,

which simplifies to

$x=\sqrt{x+n}$ .

Squaring both sides and solving for $x$ , we obtain

$x=\frac{1+\sqrt{4n+1}}{2}$ .

In order for $x$ to be an integer, our discriminant (in this case, $4n+1$ ) must be an odd perfect square. Because $n \leq 999$ , we have that $4n+1\leq 3997$ . The largest odd square less than $3997$ is $3969$ . Therefore, we have

$4n+1=3997$ ,

which implies that $n=992$ .

let a = the infinite radical series. now square it and we see a^2 - a - n = 0 ; so the solution is of the form (1+sqrt(1+4n))/2=a ; for an integer a 1+4n=y^2 since it has to be an integer value; now use case checking by starting with greatest case of n (999) and solve for y. We see that for n=999 y=63.3225644....... so 63 is highest y value. so (63^2 -1)/4=n(max) ; n(max)=992

just find out the product of the greatest consecutive no. less than 999

(expression) = K = (expression)² - n >> K = K² - n

K>0: K = 1/2[1 + (1+4n)^(1/2)]

1+4n = L²

for n=999 >> L = 63.22183.... >> L<= 63 for L=63 >> n=992 (answer)

Let $A = \sqrt{n + \sqrt{n + \ldots}}$ ; thus, $A = \sqrt{n + A}$ . We can solve for $A$ in terms of $n$ by squaring both sides and rearranging to get $A^2 - A - n = 0$ , whose solution is given by the quadratic formula as $A = \frac{1 + \sqrt{1 + 4n}}{2}.$ For $A$ to be an integer, therefore, $4n + 1$ must be an odd square; the restriction $n \leq 999$ means that $4n+1$ must be the greatest odd square less than $3997$ , or $63^2 = 3969$ . $4n + 1 = 3969$ is easily solved to yield $n = 992$ .

$x=\sqrt{n+\sqrt{n+\sqrt{n+...}}}$ implies $x^{2}=n+\sqrt{n+\sqrt{n+...}}$ , which implies $x^{2}-n=\sqrt{n+\sqrt{n+...}}=x$ , or rewritten, $x^{2}-x-n=0$ . By the Quadratic Formula, we have the following:

$x=\frac{1+\sqrt{1+4n}}{2}$ (Note: we cannot have the negative root because x is positive)

We then search for the largest value of $n$ that will make $x$ an integer. Clearly, since 1 is odd, $\sqrt{1+4n}$ needs to be odd to make $x$ an integer, which it will be since $1+4n$ is odd.

Thus, we still need $1+4n$ to be a perfect square, say $k^{2}$ . We then set $k= [\sqrt{1+4 \times 999}]=63$ and then we solve for $n$ as $n=\frac{k^{2}-1}{4}=992$ .

An infinitely nested radical of the above mentioned form can be calculated using the expression $\frac{1}{2}(1+\sqrt{1+4n})$ , where $n$ is the integer we are using in the nested radical.

Now, let us try this equation with the biggest integer allowed for this problem, which is $999$ .

$\frac{1}{2}(1+\sqrt{1+4\cdot 999})=\frac{1}{2}(1+63,2218)$

It is obvious that the integer $999$ does not satisfy our needs of another integer. However, it does show us that the radical inside the equation is the one causing us trouble. It also shows us that the nearest perfect square to $3997$ $(1+4\cdot999)$ , is $3969$ , which is actually $63^{2}$ . Therefore, if we are looking for the largest positive integer that is a solution to the infinitely nested radical, it is obvious that

$\sqrt{1+4n}=63\Rightarrow 1+4n=3969\Rightarrow 4n=3968\Rightarrow n=992$ .

We may also continue the expression just to prove the validity of this

$\frac{1}{2}(1+63)=32$ , which is indeed a positive integer, meaning that $n=992$ is a solution to the problem.

Let the expression in the question equal $x$ . If we square $x$ and subtract $n$ from that result, we find that the final result is equal to x. This gives $x^2-n=x$ . We can simplify this to $x^2-x-n$ . Any quadratic equation in the form of $x^2-x-n$ which is factorable (will give one positive integer solution) will be of the form $(x+a) \times (x-(a-1))$ where $a\times(a-1)=n$ . Since we want $n$ to be as close to $999$ as possible, a will be somewhere near the square root of $999$ (around $31.6$ ). If we try $31\times32$ , we get $992$ which is very close. If we try $32\times33$ , we get an amount over $999$ . This means that the answer must be $\boxed{992}$ .

Let $\sqrt{n+\sqrt{n+\sqrt{n}+...}}=a$ . $\sqrt{n+a}=a$ . $n+a=a^2, a^2-a-n=0$ . By quadratic formula, $a=\frac{1+\sqrt{4n+1}}{2}$ . We reject the negative sign since $a$ is always positive. $\sqrt{4n+1}$ is an odd positive integer, $4n+1$ is an odd perfect square. Since $4n+1$ is always odd, let $4n+1=b^2$ . $n=\frac{b^2-1}{4}$ . Since $n\leq999$ , $\frac{b^2-1}{4}\leq999$ . $b^2\leq3997$ . Since $b^2\leq3997<4069=64^2$ and $3997>3969=63^2$ , $b_{max}=63$ . $n_{max}=\frac{63^2-1}{4}=992$ .

I have to write this in text; I don't have the time to learn the formatting. Sorry.

Note that the expression contains itself. If we say that x = (the expression here), then I get x = sqrt(n + x). Square both sides and get x^2 = x + n. At this point I just fiddled around looking for a value of X that would work. 32^2 is slightly more than our upper bound of 1000, at 1024. 1024 - 32 = 992, so for n=992 the infinitely nested radical must evaluate to 32.

When we are given an infinitely nested radical, it is usually good to assign a variable for it.

Let us call this $x$ .

If we square it we get $x^{2}=x+n$ .

Manipulating it we get $n=x^{2}-x$ .

This factors into $n=x(x-1)$

Notice that $n$ cannot be more than $1000$ .

Also we know that $32^{2}=1024$

So the largest value of $n=1024-32=992$

Let us consider \sqrt{n+\sqrt{n+\sqrt{n+\ldots}}} = k for some positive integer k. We need to find the largest integer n\leq1000 such that the above equation is satisfied. \sqrt{n+\sqrt{n+\sqrt{n+\ldots}}}=k is equivalently \sqrt{n+k}=k since it is infinitely nested \sqrt{n+k}=k squaring on both sides n+k=k^2 n=k^2-k n=k(k-1) The largest number n less than thousand that can be written as n=k(k-1) for some positive integer k is 992= 32(31) Hence n=992 is the solution

If $\sqrt{n+\sqrt{n+\cdots}}=N\in\mathbb{Z}$ , then $N^2=n+\sqrt{n+\sqrt{n+\cdots}}=n+N$ . Solving this quadratic equation, we obtain $N=\frac{1}{2}(1+\sqrt{4n+1})$ . In order for $N$ to be integer, the expression under the root must be a perfect square of some odd number $m$ .

So, $4n+1=m^2$ or $(m-1)(m+1) \equiv 0 \pmod{4}$ , which is generally true for each odd $m$ as in that case the first multiplicand is divisible by 2 and so is the second one.

Taking into account that $m^2-1=4n<4000$ , $m<\sqrt{4001}=63.2\ldots$ . The greatest number $m$ we get thus is 63, and corresponding $n = (m^2-1)/4=992$ .

Se √(n+√(n+√(n+√(n+⋯)) ) ) =k logo (√(n+√(n+√(n+√(n+⋯)) ) ) )^2=k^2. Teremos n+√(n+√(n+√(n+⋯)) ) =k^2 então n+k=k^2. Resolvendo a equação do 2º grau para k positivo teremos k=(1+√(1+4n))/2 . Se n = 999, √(1+4*999)=√3997 Entretanto 3997 não é quadrado perfeito. Sabemos que 〖63〗^2=3969 e 〖64〗^2=4096. O quadrado perfeito menor e mais próximo de 3997 é o número 3969. Assim 1+4n=3969=>n=992.

Assume $a = \sqrt{n+\sqrt{n + \sqrt{n + ...}}}$ is the largest integer such that n is integer and n <= 999. It turn out $a^{2} = n + a$ So, the larger root of the equation $a^{2} - a - n = 0$ must be as large as possible (n <= 999). More precisely, we will have $a = \frac{1 + \sqrt{1+4n}}{2} <= \frac{1 + \sqrt{1+4*999}}{2} < \frac{1 + \sqrt{64^{2}}}{2}$ And then we can easily figure out n = 992 (with the fact that $\sqrt{1+4n}$ is the largest odd integer)

We can write: $x = \sqrt{n + \sqrt{n + \sqrt{n + \dots}}} = \sqrt{n + x}$ Squaring both sides and rearranging, we get the quadratic $x^2 - x - n = 0$ When we solve for x, we get $x = \dfrac{1 + \sqrt{(1 + 4n)}}{2}$ Plugging in n = 999, we get $x \approx 63$ . To find the largest integer solution less than 999, we calculate $\dfrac{63^2 - 1}{4}$ , which gives 992, our final answer.

Let S be the equation. Observe that $S^{2}$ = n + S. So, $S^{2}$ - S -n = 0. --- (1). Note that (1) has positive integer roots less or equal to 999. Hence, 992 = 32 x 31 is the largest positive n.

That was a interesting question, the 992 is the biggest one in n<999 that can make a integer or a half, this one make a half so it's the awnser

Let the infinitely nested radical equals x

n+x=x^2

x^2-x-n=0

n<=999<33x32

so n=32x31=992

A=√n+√n+√n+... -> A^2= n+ √n+√n+√n+... -> A^2 = n + A ->A^2 -A -n=0 resolvendo a função o maior valor possível é 992

0.5(1 + Sqrt[1 + 4n]) is a formula derived from the infinitely nested radical expression.

To solve for the largest number, under 1000, I took out the square root part and tested numbers starting from 999 and decreasing, to make sure it equaled a positive integer.

Sqrt[1 + 4n]

The one that works (992) is the answer to this question.

Let $sqrt{n+\sqrt{n+...}}$ be x. Therefore x^{2} = n + \(sqrt{n+\sqrt{n+...}} So, $x^{2}$ -x = n implies x(x-1) = n Therefore, n must be a multiple of two consecutive numbers that is lesser than and nearest to 999 . The value of n hence is 992.

1+4n is a perfect square and n<=999... Max value of n = 992