TITLE: Show that if $a+b+c=0$, $2(a^4 + b^4+ c^4)$ is a perfect square
QUESTION [6 upvotes]: 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.

REPLY [2 votes]: Denote:
$$a+b+c=0; ab+ac+bc=k; abc=t$$
Then $a,b,c$ are the roots of:
$$x^3+kx+t=0$$
Note:
$$a^3+ka+t=0 \Rightarrow a^4+ka^2+ta=0,$$
$$b^3+kb+t=0 \Rightarrow b^4+kb^2+tb=0,$$
$$c^3+kc+t=0 \Rightarrow c^4+kc^2+tc=0.$$
Add and multiply by $2$:
$$2(a^4+b^4+c^4)=-2k(a^2+b^2+c^2)-t(a+b+c)=-2k((a+b+c)^2-2k)-0=(2k)^2.$$