2. According to the condition, we write the equation where $x$ is the given number: $x^{2}+15=y^{2}$, $y$ is an integer. We transform the equation and solve it in integers. $\mathrm{y}^{2}-\mathrm{x}^{2}=15$, $(\mathrm{y}-\mathrm{x})(\mathrm{y}+\mathrm{x})=15$. From this, we obtain eight systems of equations:

1), 2) $y-x= \pm 1$, $y+x= \pm 15$; 3), 4) $y-x= \pm 15, y+x= \pm 1$; 5), 6) $y-x= \pm 3, y+x= \pm 5$; 7), 8) $y-x= \pm 5, y+x= \pm 3$

The solutions to these systems are the pairs: $(7,8),(-7,-8),(-7,8),(7,-8),(1,4),(-1,-4)$, $(-1,4),(1,-4)$.
Answer: there are such numbers: $7, -7, 1, -1$.