Find two consecutive negative integers, sum of whose squares is 481.

Let two consecutive negative integers be x and (x + $\displaystyle 1$) According to given statement, \[\begin{aligned} & x^{2}+(x+1)^{2}=481 \\ \Rightarrow & x^{2}+x-240=0 \\ \Rightarrow & (x+16)(x-15)=0 \\ \therefore & x=-16 \text { or } 15 \end{aligned} \] \(\displaystyle \mathrm{x}=15\) does not satisfy the given condition. So, required integers are - $\displaystyle 16$ and - 15.