## Task B-4.1.

In which number system does the equality $\sqrt{2521}-\sqrt{2400}=1$ hold?
## Solution.

Let the base of the sought number system be denoted by $b$. Clearly, $b>5$.

$$
\begin{aligned}
& \sqrt{2521}-\sqrt{2400}=1 \\
& \Rightarrow \sqrt{2521}=1+\sqrt{2400} /^{2} \\
& \Rightarrow 2521-1-2400=2 \sqrt{2400} \\
& \Rightarrow 120=2 \sqrt{2400}
\end{aligned}
$$

This equality, written in the system with base $b$, is

$$
\begin{array}{ll}
b^{2}+2 b=2 \sqrt{2 b^{3}+4 b^{2}} /^{2} & 2 \text { points } \\
\Rightarrow b^{4}-4 b^{3}-12 b^{2}=0 & \\
\Rightarrow b^{2} \cdot\left(b^{2}-4 b-12\right)=0 . & 2 \text { points }
\end{array}
$$

The solutions to this equation are $b_{1}=0, b_{2}=-2, b_{3}=6$.

The base of the sought number system is $b=6$.