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.
 A: 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.

†
$$(a^2+b^2+c^2)^3\geqslant \overbrace{a^6+b^6+c^6+\underbrace{a^4b^2+a^2b^4}_\text{apply AM-GM}+\underbrace{b^4c^2+b^2c^4}_\text{here too}+\underbrace{a^4c^2+a^2c^4}_\text{and here}}^\text{skipped quite a lot of terms, all of them quadratic}\geqslant \\\geqslant a^6+b^6+c^6+2a^3b^3+2b^3c^3+2a^3c^3=(a^3+b^3+c^3)^2,\text{ hence}\\
a^3+b^3+c^3\leqslant (a^2+b^2+c^2)^{3/2}=4^{3/2}=8$$
A: A somewhat simpler solution is given by Lagrange multipliers. The stationary points of $f(a,b,c)=a^3+b^3+c^3$ over $a^2+b^2+c^2=4$ occur where $3(a^2,b^2,c^2)=2\lambda(a,b,c)$, i.e. at $\pm\left(\frac{2}{\sqrt{3}},\frac{2}{\sqrt{3}},\frac{2}{\sqrt{3}}\right)$ and at the permutations of $\pm(0,0,2)$ and $\pm(0,\sqrt{2},\sqrt{2})$. In particular
$$ \max_{a^2+b^2+c^2}a^3+b^3+c^3 = 8 $$
and in a point of absolute maximum we have $a^4+b^4+c^4=16$.
