Đặt \(a=2015,\) ta có
\(a^2+a^2\cdot\left(a+1\right)^2+\left(a+1\right)^2=a^2\left(1+\left(a+1\right)^2\right)+\left(a+1\right)^2=a^2\left(a^2+2a+2\right)+\left(a+1\right)^2\)
\(=2a^2\left(a^2+a+1\right)-a^4+\left(a+1\right)^2=2a^2\left(a^2+a+1\right)-\left(a^2-a-1\right)\left(a^2+a+1\right)\)
\(=\left(a^2+a+1\right)\left(2a^2-a^2+a+1\right)=\left(a^2+a+1\right)^2.\)
Vậy ta được \(\sqrt{2015^2+2015^2\cdot2016^2+2016^2}=a^2+a+1\) là số nguyên dương. (ĐPCM)