Three consecutive integers are such that sum of the square of second and product of other two is $\displaystyle 161$ . Find the three integers.

Let the three numbers be \(\displaystyle \mathrm{x}, \mathrm{x}+1\) and \(\displaystyle \mathrm{x}+2\) \[\begin{aligned} & \Rightarrow(\mathrm{x}+1)^{2}+\mathrm{x}(\mathrm{x}+2)=161 \\ & \Rightarrow \mathrm{x}^{2}+2 \mathrm{x}-80=0 \\ & \Rightarrow(\mathrm{x}+10)(\mathrm{x}-8)=0 \\ & \therefore \mathrm{x}=8 \text { or }-10 \end{aligned} \] So, the numbers are $\displaystyle 8$, $\displaystyle 9$, $\displaystyle 10$ or -$\displaystyle 10$, -$\displaystyle 9$, -$\displaystyle 8$