Câu hỏi 79
Tìm tất cả số nguyên dương $n$ sao cho
$$\lfloor\sqrt[4]{1}\rfloor+\lfloor\sqrt[4]{2}\rfloor+\dots+\lfloor\sqrt[4]{n}\rfloor=\frac32n+1.$$
$$\lfloor\sqrt[4]{1}\rfloor+\lfloor\sqrt[4]{2}\rfloor+\dots+\lfloor\sqrt[4]{n}\rfloor=\frac32n+1.$$
Math Olympiad Vault · MathNet 2013