Square Roots Rooted

Let's put $\sqrt{\sqrt{x^2+1000x}-x}=k$ where $k$ is a positive integer.

Square it:

$\sqrt{x^2+1000x}-x=k^2$ .

Rearrange:

$\sqrt{x^2+1000x}=k^2+x$

Square again:

$x^2+1000x=(k^2+x)^2$

Rearrange:

$1000x=(k^2+x)^2-x^2=(k^2+x+x)(k^2+x-x)=(k^2+2x)\cdot k^2=k^4+2xk^2$

$\Rightarrow x=\frac{k^4}{1000-2k^2}$ .

Now we have to look for positive integers $k$ so that $\frac{k^4}{1000-2k^2}$ is a positive integer.

Notice that if $k\geq 23$ , $x$ becomes negative. So, $k<23$ .

Also, notice that $k>4$ . Otherwise the value of the denominator will be greater than the value of the numerator making $x$ a proper faction [not an integer].

Finally notice that the denominator is even. So, the numerator also has to be even. So, $2 | k^4$ . That means $2 | k$ .

So, we can conclude that $k$ is an even number.

So far we've narrowed $k$ down to an even number in the range $4<k<23$ . So, we're left with $9$ values of $k$ . Unfortunately, I can't do any better.

So, we're going to check cases by plugging in all the possible values of $k$ . If we do that, we'll see that $k=20$ is the only value that makes $x$ a positive integer. And when $k=20$ , $x=\frac{20^4}{1000-2\times 20^2}=800$ .

Since the only possible value of $x$ is $800$ , the sum of all possible values of $x$ is simply $800$ .

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My solution isn't that good. Because in the end, I had to go through a lot of tedious calculations. There has to be a better way than that. And I'm looking forward to better solutions.

Let $z = \sqrt{\sqrt{x^2+1000x}-x}$ . Assume that $z$ is a nonnegative integer, as is desired in the conditions of the problem. Then, $x = \frac{z^4}{1000-2z^2}$ . Also, clearly, $z \neq 0$ because then we get $x=0$ , which is not a positive integer. Additionally, since $(1000-2z^2) | z^4$ , we have that $2|z^4$ and thus $2|z$ .

Since $x$ is a positive integer, we have that $z^4 \ge 1000-2z^2$ , or that $z \ge 6$ .

We also get the inequality $1000 \ge 2z^2$ , or that $z \le 22.$ Observe that $\frac{z^4}{1000-2z^2}$ is an integer iff $\frac{z^4}{1000-2z^2}+\frac{1}{2}z^2+250=\frac{250000}{1000-2z^2}$ is an integer. Thus, $(1000-2z^2) | 250000$ , implying that $1000-2z^2$ can only have the prime factors 2 and 5. It is clear, then, that $z$ can only have the prime factors 2 and 5. Given these conditions, $z$ can only be $20, 16, 10$ or $8$ . Checking these values reveals that only $z=20$ works, which gives $x = \boxed{800}$ .

Let m = sqrt(sqrt(x^2 +1000x) - x). Then (m^2 + x )^2 = x^2 + 1000x, which simplifies to

(1000 – 2m^2)x = m^4.

Note here that the left hand side is even. So m^4 is even. So m = 2k, where k ∈ ℤ+. Thus

(1000 – 8k^2)x = 16k^4,

which simplifies to

(*): x = 2k^4 / (125 – k^2).

Now x ∈ ℤ+. So since 2k^4 > 0, we must have 125 – k^2 > 0. This means that

{ k ∈ ℤ+ | 2k^4 / (125 – k^2) ∈ ℤ+ } ⊆ { 1, ..., 11}.

A quick inspection of (*) with k ∈ {1, ..., 11} leads to k = 10 as the only value of k that makes x ∈ ℤ+. So the answer is 2⋅10^4 / (125 – 10^2) = 800.

x^2 + 1000x = k^2 that is ... (x+500)^2-500^2=k^2 or (x+k+500)(k-x+500)=250000 let k+x=m and k-x=n

then (m+500)(500-n)=250000=5^6into2^4 now the expression in question can be concised into (k-x)^0.5

thus we just need to find all those value of (k-x) which are squares and (k-x) or (n) is nothing but 500-f

where f is the factor of 25000 such that it is always smaller then 500.

we get following cases of f=16,100,400 as 500-16=484=22into22 and 500-100=400=20into20

500-400=100=10into10. now for f=16 or m+500=5^6

which is odd hence we dont get integral solution for m . similarly when f=400 500+m=5^4 which is again odd.

hence the only case to consider if f=100 then 500+m=2500 or m=2000 or k+x=2000 and k-x=400

solving above equation we get x=800 which is the required solution.

Let $\sqrt{\sqrt{x^{2}+1000x}-x}=k,$ for some positive integer $k.$ Notice that $x^{2} < x^{2}+1000x < x^{2}+1000x+250000$ $\iff x < \sqrt{x^{2}+1000x} < x+500$ So, we can substitute $(x^{2}+1000x)$ by $(x+p)^2$ for some integer $p\in{(0,500)}.$ Thus, $\sqrt{\sqrt{x^{2}+1000x}-x}=k\iff\sqrt{\sqrt{(x+p)^2}-x}=k$ $\iff \sqrt{(x+p)-x}=k$ $\iff \sqrt{p}=k$ From the last equation, we get $p$ is a square number less than $500$ But, $x^{2}+1000x=(x+p)^2$ $\iff x^{2}+1000x=x^2+2px+p^2$ $\iff2(500-p)x=p^2$ Since $2|2(500-p)x,$ then $2|p^2$ So, all possible values of $p$ are $4,16,36,64,100,144,196,256,324,400,484.$ For $p\leq36,$ $2(500-p)>p^2$ , which causes $x$ is not integer number. Checking the rest of all possible values of $p,$ we get $p=400$ is the only value that satisfies the problem. So, $2(500-p)x=p^2$ $\iff 200x=1600$ $\iff x=800$ Substituting $x=800$ to the equation, we get $\sqrt{\sqrt{800^{2}+1000 \cdot 800}-800}$ $=\sqrt{\sqrt{1440000}-800}$ $=\sqrt{400}$ $=20,$ which is an integer number. Hence the sum of all possible value(s) of $x$ is $800$ .

Assume any variable y equivalent to the above expression. This results in x = y^4 - (1000 - 2y^2). Now, solve for positive values of x. For y = 20, x = 800. That's your answer.

VBA: code

For x = 1 To 1000000

p = Sqr(Sqr((x ^ 2 + 1000 * x)) - x)

If p = Int(p) Then

s = s + x

End If

Next x

MsgBox (s)

Output:800