## Task A-4.1.

Determine all natural numbers $n$ for which

$$
\lfloor\sqrt[4]{1}\rfloor+\lfloor\sqrt[4]{2}\rfloor+\ldots+\lfloor\sqrt[4]{n}\rfloor=\frac{3}{2} n+1
$$

( $\lfloor x\rfloor$ denotes the greatest integer not greater than $x$.)
## Solution.

Let

$$
a_{n}=\lfloor\sqrt[4]{1}\rfloor+\lfloor\sqrt[4]{2}\rfloor+\ldots+\lfloor\sqrt[4]{n}\rfloor
$$

We need to solve the equation

$$
a_{n}=\frac{3}{2} n+1
$$

Note that $\lfloor\sqrt[4]{k}\rfloor=1$ for $1 \leqslant k<16$.

Therefore, for $1 \leqslant n \leqslant 15$, we have $a_{n}=n$.

Among these numbers, there are no numbers for which the given equality holds.

The equality $\lfloor\sqrt[4]{k}\rfloor=2$ holds for $16 \leqslant k<81$.

Therefore, for $16 \leqslant n \leqslant 80$, we have $a_{n}=1+1+\ldots+1+2+\ldots+2=15+2 \cdot(n-15)=2 n-15$. 1 point

For $16 \leqslant n \leqslant 80$ to satisfy equation $(*)$, it must be that $2 n-15=\frac{3}{2} n+1$.

We get one solution $n=32$, and this is the only solution of the problem less than 81.

1 point

There are no other (larger) solutions, because after $n=81$, with each increase of $n$ by 1, the value of the left side increases by at least 3, while the value on the right side increases by $\frac{3}{2}$, so the left side will remain greater than the right side.