Let

f(x) = \sqrt{x - \sqrt{x}}

Find the largest three-digit value of x such that f(x) is an integer.

parmen May 1, 2024

#1**0 **

Suppose f(x) is an integer. Consequently sqrt(x) is an integer. Suppose N = sqrt(x). Then sqrt(x - sqrt(x)) = sqrt(N^2 - N) = sqrt(N(N-1)).

x is 3 digits, so the range of N is \(\sqrt{100} \leq N \leq \sqrt{961} \), which is \(10 \leq N\leq 31\). There are no integer in this range where N(N - 1) is a square number.

Hence, there are no solution.

MaxWong May 1, 2024