﻿ Compute the smallest positive integer n for which \sqrt{100+\sqrt{n} } +\sqrt{100-\sqrt{n} } is an integer

#### Question

Compute the smallest positive integer n for which

\sqrt{100+\sqrt{n} } +\sqrt{100-\sqrt{n} }

is an integer

Collected in the board: Cube root

Steven Zheng posted 1 year ago

Let

a = \sqrt{100+\sqrt{n} }
(1)

b = \sqrt{100-\sqrt{n} }
(2)

and

k = \sqrt{100+\sqrt{n} } +\sqrt{100-\sqrt{n} }

From the given condition, we know k is an integer and,

a+b = k
(3)
ab = \sqrt{100+\sqrt{n} }\cdotp \sqrt{100-\sqrt{n} }

=\sqrt{10000-n}

Rearrange the above equation, we get

n = 10000-(ab)^2
(4)

Square both (1) and (2), and addition of squared equations

a^2+b^2 = 200
(5)

Square (3) and subtract (5)

2ab = k^2-200

Then

ab = \dfrac{k^2}{2}-100
(6)

Since ab>0, \dfrac{k^2}{2}-100>0, then k>10\sqrt{2}

Substituting (6) to (4) gives

n = 10000-(\dfrac{k^2}{2}-100)^2
(7)

Subtracting 2 times (6) from (5) gives

a^2+b^2-2ab = 200-k^2+200

(a-b)^2 = 400-k^2

Since a > b, we can cancel negative result when taking square root.

a-b = \sqrt{400-k^2}
(8)

We can solve the equation system formed by (3) and (8) to determine a and b. But here we don't need to do so to answer the question.

In order for the equation (8) to be meaningful, the radicant must be greater than 0, that is

400-k^2>0

Solving the inequality gives

k<20

Since equation (7) is a decreasing function when k>10\sqrt{2}, the smallest n happens when k is the largest number. Exclude k=19 since it yields non integer for n. Therefore,

k = 18

Substituting (7) yields,

n = 10000-(\dfrac{18^2}{2} -100)^2 =10000-62^2 = 6156

Steven Zheng posted 1 year ago

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