[Algebra]
Problem: Compute the smallest positive integer $n$ such that

$$
n,\lfloor\sqrt{n}\rfloor,\lfloor\sqrt[3]{n}\rfloor,\lfloor\sqrt[4]{n}\rfloor,\lfloor\sqrt[5]{n}\rfloor,\lfloor\sqrt[6]{n}\rfloor,\lfloor\sqrt[7]{n}\rfloor, \text { and }\lfloor\sqrt[8]{n}\rfloor
$$

are distinct.

Solution: Inverting the problem, the goal is to find seven positive integers $a<b<c<d<e<f<g$ and a positive integer $n$ such that $a^{8}, b^{7}, c^{6}, \ldots, g^{2} \leq n$ and $n<(a+1)^{8},(b+1)^{7}, \ldots,(g+1)^{2}$. Proceed by cases starting with small values of $a$.

If $a=1$, then because $n<(a+1)^{8}, n<256$. But because $n \geq(a+3)^{5}, n \geq 4^{5}=1024$. So it is impossible for $a$ to be 1 .

If $a=2$, then $a^{8}=256$ and $(a+1)^{8}=6561$, so $256 \leq n<6561$. Then $b \geq 3 \Rightarrow b^{7} \geq 2187$ and $c \geq 4 \Rightarrow c^{6} \geq 4096$. So $n \geq 4096$. Because $(3+1)^{7}=16384$ and $(4+1)^{6}=15625$, the condition $n<6561$ found previously guarantees that $\lfloor\sqrt[7]{n}\rfloor=3$ and $\lfloor\sqrt[6]{n}\rfloor=4$. Notice that if $4096 \leq n<6561$, then $\lfloor\sqrt[5]{n}\rfloor=5,\lfloor\sqrt[4]{n}\rfloor=8$, and $\lfloor\sqrt[3]{n}\rfloor \geq 16$. In fact, $\lfloor\sqrt[3]{4096}\rfloor=2^{4}=16$ and $\lfloor\sqrt{4096}\rfloor=2^{6}=64$. So the desired value of $n$ is 4096 .

Final answer: 4096