If $a^2 + b^2 +c^2 =4$ and $a^3 + b^3 + c^3 = 8$ then find $a^4+b^4+c^4$ This is a popular question, but I can find $a^4+b^4+c^4$ if I know $a+b+c, a^2+b^2+c^2, a^3 +b^3 +c^3$ or in special case $a+b+c=0$ we only need a more condition like $a+b+c$. I have tried in many times and can't find out the way to archive it.

Well, it sounds logical that we need three equations, because three variables and stuff. This is wrong, though. Sometimes one equation can give us two variables, like $x^2+y^2=0\Rightarrow x=0,\;y=0$ (if we stay in $\mathbb R$, that is; but if we don't, then the original question can't be answered either). Same thing here.
Obviously†, $a^3+b^3+c^3\leqslant(a^2+b^2+c^2)^{3/2}$, and the equality is only possible when two of the variables are $0$. So $(0,0,2)$ and its permutations are indeed the only real solutions, from where you can easily find the answer.
So it goes.

†
$$(a^2+b^2+c^2)^3\geqslant \overbrace{a^6+b^6+c^6+\underbrace{a^4b^2+a^2b^4}_\text{apply AM-GM}+\underbrace{b^4c^2+b^2c^4}_\text{here too}+\underbrace{a^4c^2+a^2c^4}_\text{and here}}^\text{skipped quite a lot of terms, all of them quadratic}\geqslant \\\geqslant a^6+b^6+c^6+2a^3b^3+2b^3c^3+2a^3c^3=(a^3+b^3+c^3)^2,\text{ hence}\\
a^3+b^3+c^3\leqslant (a^2+b^2+c^2)^{3/2}=4^{3/2}=8$$
