<s>[INST] Show that if $a+b+c=0$, $2(a^4 + b^4+ c^4)$ is a perfect square 
Show that for $\{a,b,c\}\subset\Bbb Z$ if $a+b+c=0$ then $2(a^4 + b^4+ c^4)$ is a perfect square. 

This question is from a math olympiad contest. 
I started developing the expression $(a^2+b^2+c^2)^2=a^4+b^4+c^4+2a^2b^2+2a^2c^2+2b^2c^2$ but was not able to find any useful direction after that.
Note: After getting 6 answers here, another user pointed out other question in the site with similar but not identical content (see above), but the 7 answers presented include more comprehensive approaches to similar problems (e.g. newton identities and other methods) that I found more useful, as compared with the 3 answers provided to the other question. 
 [/INST] A systematic way doing this is using Newton's identifites.
Let $p_k = a^k + b^k + c^k$ for $k = 1, 2, 3, 4$ and
$$\begin{align}
s_1 &= a + b + c\\
s_2 &= ab+bc+ca\\
s_3 &= abc
\end{align}$$ 
be the elementary symmetric polynomials associated with $a, b, c$.
Newton's identities tell us:
$$\require{cancel}\begin{array}{rlrlrlrlrl}
p_1 &-& s_1 &&&&&= 0\\
p_2 &-& 
\cancelto{ 0}{\color{grey}{s_1 p_1}}
&+& 2s_2 &&&= 0\\
p_3 &-& 
\cancelto{ 0}{\color{grey}{s_1 p_2}}
&+& 
\cancelto{ 0}{\color{grey}{s_2 p_1}}
&- &3s_3 &= 0\\
p_4 &-& 
\cancelto{ 0}{\color{grey}{s_1 p_3}}
&+& s_2 p_2 &-&
\cancelto{ 0}{\color{black}{s_3 p_1}}
&= 0
\end{array}
$$
When $a + b + c = 0$, $s_1 = 0$ and $1^{st}$ equation $p_1 - s_1 = 0$ tell us $p_1 = 0$.
Substitute back into $2^{nd}$ and $4^{th}$ equations lead to
$$\begin{cases}
p_2 = -2s_2,\\
p_4 = -s_2 p_2
\end{cases}
\quad\implies\quad
2(a^4+b^4+c^4) = 2p_4 = -2s_2 p_2 = p_2^2 = (a^2+b^2+c^2)^2$$
</s>