Two consecutive even positive integers, sum of the squares of which is 1060, are · ABC26GN0410

Let, the two consecutive even numbers are $a$ and $(a+2)$, respectively According to question, $$\begin{aligned} & a^{2}+(a+2)^{2}=1060\left\{\because(a+b)^{2}=a^{2}+b^{2}+2 a b\right\} \\ & \Rightarrow a^{2}+a^{2}+4+4 a=1060 \\ & \Rightarrow 2 a^{2}+4 a-1056=0 \\ & \Rightarrow a^{2}+2 a-528=0 \\ & \Rightarrow a^{2}+24 a-22 a-528=0 \\ & \Rightarrow a(a+24)-22(a+24)=0 \\ & \Rightarrow(a-22)(a+24)=0 \\ & \Rightarrow a=-24,22 \end{aligned}$$