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. 

Since $$2(a^2b^2+a^2c^2+b^2c^2)-a^4-b^4-c^4=(a+b+c)(a+b-c)(a+c-b)(b+c-a)=0,$$
we obtain
$$2(a^4+b^4+c^4)=a^4+b^4+c^4+2(a^2b^2+a^2c^2+b^2c^2)=(a^2+b^2+c^2)^2.$$
Done!
