Giải thích các bước giải:
Ta có:
$A=2012^{4n}+2013^{4n}+2014^{4n}+2015^{4n}(1)$ chẵn vì $n\in Z^+$
Mà:
$2012\equiv 0(mod 4)\to 2012^{4n}\equiv 0(mod 4)$
$2013\equiv 1(mod 4)\to 2013^{4n}\equiv 1(mod 4)$
$2014\equiv 0(mod 2)\to 2014^2\equiv 0(mod 4)\to (2014^2)^{2n}\equiv 0(mod 4)\to 2014^{4n}\equiv 0(mod 4)$
$2015\equiv -1(mod 4)\to 2015^{4n}\equiv 1(mod 4)$
$\to 2012^{4n}+2013^{4n}+2014^{4n}+2015^{4n}\equiv 0+1+0+1\equiv 2(mod 4)$
$\to 2012^{4n}+2013^{4n}+2014^{4n}+2015^{4n}\quad\not\vdots\quad 4(2)$
Từ $(1),(2)\to A$ không là số chính phương
