Show that if $a+b+c=0$, $2(a^4 + b^4+ c^4)$ is a perfect square

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.

$$c^4=(-a-b)^4=(a+b)^4=a^4+4a^3b+6a^2b^2+4ab^3+b^4$$

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 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

$$\begin{cases} p_2 = -2s_2,\\ p_4 = -s_2 p_2 \end{cases} \quad\implies\quad 2(a^4+b^4+c^4) = 2p_4 = -2s_2 p_2 = p_2^2 = (a^2+b^2+c^2)^2$$