$ $
풀이. $c = -a-b$이므로,
\[ \begin{align*}
\frac{a^4 + b^4 + c^4}{2}
&= \frac{1}{2} ( a^4 + b^4 + (-a-b)^4 ) \\[5px]
&= \frac{1}{2} ( 2a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + 2b^4 ) \\[5px]
&= a^4 + \textcolor{red}{2a^3b} + \textcolor{blue}{3a^2b^2} + \textcolor{green}{2ab^3} + b^4 \\[5px]
&= a^4 + \textcolor{red}{a^3b} + \textcolor{blue}{a^2b^2} + \textcolor{red}{a^3b} + \textcolor{blue}{a^2b^2} + \textcolor{green}{ab^3} + \textcolor{blue}{a^2b^2} + \textcolor{green}{ab^3} + b^4 \\[5px]
&= a^2(a^2 + ab + b^2) + ab(a^2 + ab + b^2) + b^2(a^2 + ab + b^2) \\[5px]
&= (a^2 + ab + b^2)^2
\end{align*} \]
따라서 좌변은 언제나 완전제곱수이다.$ $
$ $
참고. 간단한 계산을 통해서 실제로
\[ \frac{a^4 + b^4 + c^4}{2} = (a^2 + ab + b^2)^2 = (b^2 + bc + c^2)^2 = (c^2 + ca + a^2)^2 = (ab + bc + ca)^2 \]
가 모두 성립함을 확인할 수 있다.