23. [12] Compute the smallest positive integer $n$ for which
$$
0<\sqrt[4]{n}-\lfloor\sqrt[4]{n}\rfloor<\frac{1}{2015} .
$$
Answer: 4097
Let $n=a^{4}+b$ where $a, b$ are integers and $01$, so $a=8$ is the minimum.
Hence, the minimum $n$ is $8^{4}+1=4097$.