Power sums from the first three

Idea. Compute $e_1,e_2,e_3$, then use the polynomial satisfied by each root to push the power sums up one degree at a time.

Step 1: $e_2$. From $(a+b+c)^2=a^2+b^2+c^2+2e_2$ we get $4=6+2e_2$, so $e_2=-1$.

Step 2: $e_3$. Each root satisfies $x^3=2x^2+x+e_3$ (since $e_1=2$, $e_2=-1$). Summing over $x=a,b,c$:
$$8=p_3=2p_2+p_1+3e_3=12+2+3e_3,\qquad\text{so } e_3=-2.$$

Step 3: $p_4$. Multiply $x^3=2x^2+x-2$ by $x$: $x^4=2x^3+x^2-2x$. Summing,
$$p_4=2p_3+p_2-2p_1=16+6-4=18.$$

Check. The polynomial is $t^3-2t^2-t+2=(t-2)(t-1)(t+1)$, with roots $2,1,-1$: indeed $p_1=2$, $p_2=6$, $p_3=8$, $p_4=16+1+1=18$.

Lesson. Newton's identities: $p_{k+3}=e_1p_{k+2}-e_2p_{k+1}+e_3p_k$. You never need the roots themselves, only the elementary symmetric polynomials.