A SPECIALLY NICE SOLUTION TO PROBLEM 4

by

Vince Wesselmann

Let a, b, c be the roots of a cubic: (x-a)(x-b)(x-c). Upon expansion, this cubic becomes x³-(a+b+c)x²+(ab+ac+bc)x-abc. We are given that a+b+c=0. Let A:=ab+bc+ac and B:=abc.

Let S_n:=aⁿ+bⁿ+cⁿ.

Clearly

$$\begin{eqnarray*}S_0 &= &3 \\ S_1& =& 0 \\ S_2 & =& a^2+b^2+c^2=(a+b+c)^2-2(ab+ac+bc)=-2A. \end{eqnarray*}$$

Claim: For $n\geq 3$, S_n=-AS_n-2+BS_n-3.

$$\begin{eqnarray*}S_n & = &a^n+b^n+c^n \\ &= &a^{n-1}(-b-c)+b^{n-1}(-a-c)+c^{n-1... ...n-2})+abc(a^{n-3}+b^{n-3}+c^{n-3}) \\ & = & -AS_{n-2}+BS_{n-3}. \end{eqnarray*}$$

Therefore

$$\begin{eqnarray*}S_3 &= &-A(0)+B(3)=3B \\ S_4 &= &-A(-2A)+B(0)=2A^2 \\ S_5 &= &-A(3B)+B(-2A)=-5AB. \end{eqnarray*}$$

Putting it all together:

$$\frac{S_3}{3}\cdot\frac{S_2}{2}=B\cdot (-A)=-AB=\frac{S_5}{5}.$$

Questions and/or comments should be directed to Judy Downey or Griff Elder

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Last modified:   Mon Feb 5 12:32:02 CST 2001