Compute the smallest possible integer $x$ such that $\lfloor \sqrt[8]{x}\rfloor < \lfloor \sqrt[7]{x}\rfloor < ... \lfloor \sqrt{x}\rfloor < x$

Question

Compute the smallest positive integer $x$ such that

$\lfloor \sqrt[8]{x}\rfloor <\lfloor \sqrt[7]{x}\rfloor <\lfloor \sqrt[6]{x}\rfloor <\lfloor \sqrt[5]{x}\rfloor <\lfloor \sqrt[4]{x}\rfloor <\lfloor \sqrt[3]{x}\rfloor <\lfloor \sqrt{x}\rfloor < x$

I do not know how to solve this problem. I tried to brute force it, but it did not work.

Answer 1

Is this a programming question? I am unsure of a mathematical trick that would yield the result, but here is a quick python script

from sympy import integer_nthroot
for x in range(100000):
    if all(integer_nthroot(x,k)[0]<integer_nthroot(x,k-1)[0] for k in range(2,9)):
        print(x)
        break

In about a second it spits out 4096 as the first integer with this property.

Answer 2

So we need $a_n = \lfloor x {\text ^} (1 / (8 - n)) \rfloor$ (for $n \in 0 \cdots 7$) to be a strictly increasing non negative sequence of integers. The smallest possible is $a = [0, 1, 2, 3, 4, 5, 6, 7]$, that is

$$a_n = \lfloor x {\text ^} (1 / (8 - n)) \rfloor \ge n$$

It at least has to hold that

$$\begin{array} {rcl} \lfloor \sqrt[8]{x} \rfloor \ge 0 & \implies & x \ge 0 \\ \lfloor \sqrt[7]{x} \rfloor \ge 1 & \implies & x \ge 1 \\ \lfloor \sqrt[6]{x} \rfloor \ge 2 & \implies & x \ge 2^6 = 64 \\ \lfloor \sqrt[5]{x} \rfloor \ge 3 & \implies & x \ge 3^5 = 243 \\ \lfloor \sqrt[4]{x} \rfloor \ge 4 & \implies & x \ge 4^4 = 256 \\ \lfloor \sqrt[3]{x} \rfloor \ge 5 & \implies & x \ge 5^3 = 125 \\ \lfloor \sqrt[2]{x} \rfloor \ge 6 & \implies & x \ge 6^2 = 36 \\ \lfloor \sqrt[1]{x} \rfloor \ge 7 & \implies & x \ge 7 \\ \end{array}$$

So $x \ge 4^4$ . But $\lfloor 4^4 \text{^} (1/8) \rfloor = \lfloor 4^4 \text{^} (1/7) \rfloor = 2$. So $a_4 \ge 5$.

The smallest $y$ satisfying $y^4 \text{^} (1/8) > y^4 \text{^} (1/7)$ is $y = 7$, so $a_4 \ge 7$.

But then $7^4 \text{^} (1/7) = 7^4 \text{^} (1/6) = 3$, so $a_4 \ge 8$.

By computation $x = 8^4$ works and so is minimal.

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This answer is intended to be an approach that would work even if it went up to large $n$ like $\sqrt[24] x < \sqrt[23] x < \dots$, where a direct enumeration would be infeasible.