1. Determine $x \in \mathbb{N}^{*}$ that satisfies the relation $[\sqrt{x}]=\frac{2015-x}{x}$, where $[a]$ is the integer part of the number $a$.
1. $[\sqrt{x}]=\frac{2015}{x}-1$ so $\frac{2015}{x} \in \mathbb{N}^{*}, x \in \mathbb{N}^{*}$ it follows that $x$ is a divisor of 2015 $2 p$

Since $2015=5 \cdot 13 \cdot 31$. By checking, it is found that among the divisors, $x=5 \cdot 31=155$ satisfies the condition. Therefore, $x=155$ $.5 p$