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. 
 A: plugging $$c=-a-b$$ in the term $$2(a^4+b^4+c^4)$$ we get $$4\, \left( {a}^{2}+ab+{b}^{2} \right) ^{2}$$ and this is a perfect square.
A: $$c^4=(-a-b)^4=(a+b)^4=a^4+4a^3b+6a^2b^2+4ab^3+b^4$$
Therefore,
$$2(a^4+b^4+c^4)=4(a^4+2a^3b+3a^2b^2+2ab^3+b^4)$$
Now compute $(a^2+ab+b^2)^2$.
A: 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
$$\begin{cases}
p_2 = -2s_2,\\
p_4 = -s_2 p_2
\end{cases}
\quad\implies\quad
2(a^4+b^4+c^4) = 2p_4 = -2s_2 p_2 = p_2^2 = (a^2+b^2+c^2)^2$$
A: $$a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)=?$$
Now $$(a^2)^2+(b^2)^2+(c^2)^2=(a^2+b^2+c^2)^2-2(a^2b^2+b^2c^2+c^2a^2)$$
$$a^2b^2+b^2c^2+c^2a^2=(ab+bc+ca)^2-abc(a+b+c)=?$$
A: 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 $2$:
$$2(a^4+b^4+c^4)=-2k(a^2+b^2+c^2)-t(a+b+c)=-2k((a+b+c)^2-2k)-0=(2k)^2.$$
A: 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!
A: We can simplify the algebra by noticing that the expressions are homogenous:
$
2(a^4+b^4+c^4)
\\\quad=2(a^4+b^4+(a+b)^4)
\\\quad=2b^4(x^4+1+(x+1)^4),
\quad x=a/b
$
$
2(x^4+1+(x+1)^4)
\\\quad=4 x^4 + 8 x^3 + 12 x^2 + 8 x + 4
\\\quad=4 (x^2 + x)^2 + 8 (x^2 + x) + 4
\\\quad=4(y^2+2y+1),
\quad y=x^2+x
\\\quad=4(y+1)^2
\\\quad=(2(y+1))^2
\\\quad=(2(x^2+x+1))^2
$
$
2(a^4+b^4+c^4)
\\\quad=b^4(2(x^2+x+1))^2
\\\quad=(2b^2(x^2+x+1))^2
\\\quad=(2(a^2+ab+b^2)^2
$
