Consider $a,b,c$ such that $a+b+c =1, a^2+b^2+c^2=2$ and $a^3+b^3+c^3=3$. Find the value of $a^4+b^4+c^4$, if possible.
Trial: I observe that \begin{align} a^4+b^4+c^4 &=(a^2+b^2+c^2)^2-2(a^2b^2+b^2c^2+c^2a^2)\\&=2^2-2[(ab+bc+ca)^2-2abc(a+b+c)]\\&=4-2[(ab+bc+ca)^2-2abc] \end{align} Then \begin{align} (a+b+c)^2 &=a^2+b^2+c^2+2(ab+bc+ca)\\\implies 1^2= 2+ 2(ab+bc+ca)\\\implies ab +bc+ca &=-\dfrac12 \end{align} Then I am stuck. Please help.
On
Hint:let $$x_{n}=a^n+b^n+c^n$$ and $$x_{n+2}=(a+b+c)x_{n+1}-(ab+bc+ac)x_{n}+abc\cdot x_{n-1}$$ so $$x_{4}=(a+b+c)x_{3}-(ab+bc+ac)x_{2}+abc\cdot x_{1}$$ and we easy to find $ab+bc+ac,abc$
On
You may just use Newton's identities. If we consider $$p(x)=(x-a)(x-b)(x-c) = x^3-e_1 x^2+e_2 x - e_3 \tag{1}$$ and define $p_n = a^n+b^n+c^n$, we have: $$\left\{\begin{array}{rcl} e_1 &=& p_1,\\ 2e_2 &=& e_1 p_1-p_2,\\3e_3&=&e_2 p_1-e_1 p_2+p_3,\tag{2}\end{array}\right.$$ hence $e_1=1, e_2 =-\frac{1}{2},e_3 = \frac{1}{6} $. Assuming $x\in\{a,b,c\}$ we have $p(x)=0$, from which: $$ x^4 = e_1 x^3 -e_2 x^2 +e_3 x \tag{3}$$ follows. Then summing $(3)$ over $x\in\{a,b,c\}$ we get: $$ p_4 = e_1 p_3 - e_2 p_2 + e_3 p_1 = 3e_1-2e_2+e_3 = 3+1+\frac{1}{6}=\color{red}{\frac{25}{6}}.\tag{4}$$ We may also notice that the discriminant of $p(x)$ is negative, hence the set $\{a,b,c\}$ is made by one real number and two (conjugated) complex numbers. That also follows from the fact that $S_1=\frac{e_1}{3},S_2=\frac{e_2}{3},S_3=e_3$ do not fulfill the Newton's inequality $S_1 S_3\leq S_2^2$.
Note that \begin{align} a^3+b^3+c^3-3abc &=(a+b+c)(a^2+b^2+c^2-ab-ac-ca)\\\implies3-3abc&=1\cdot(2-(ab+bc+ca))\\\implies abc &= \dfrac16 \end{align}