Finding $x^4 + y^4 + z^4$ using geometric series

Solution 1:

This is a rather encumbered solution for a not-as-hard problem, and it is the only way I see to make sense of what your friend is saying. On the other hand, it is cool that it let us know about Newton's identities and their proof using generating functions.

Let us form the formal series \begin{align}F(x,y,z,t)&:=(x+y+z)t+(z^2+y^2+z^2)t^2+(x^3+y^3+z^3)t^3+\ldots\\&=3t+5t^2+7t^3+at^4+\ldots,\end{align} where $a:=x^4+y^4+z^4$ is the number we want to find.

Forming this power series is kind of adding the given equations (as your friend suggested), but we multiply them first by $t,t^2,t^3,...$, respectively, before adding.

Consider the polynomial \begin{align}G(x,y,z,t)&:=-1+(x+y+z)t+(xy+xz+yz)t^2-(xyz)t^3\\&=-1+3t+(xy+xz+yz)t^2-(xyz)t^3\end{align}

We multiply $F$ times $G$. It happens that we get a polynomial:

\begin{align} FG&=\left(\sum_{k=1}^{\infty}(xt)^k+\sum_{k=1}^{\infty}(yt)^k+\sum_{k=1}^{\infty}(zt)^k\right)G\\ &=-\left(\frac{xt}{1-xt}+\frac{yt}{1-yt}+\frac{yt}{1-xt}\right)(1-xt)(1-yt)(1-zt)\\ &=t\left((-x)(1-yt)(1-zt)+(-y)(1-xt)(1-zt)+(-z)(1-xt)(1-yt)\right)\\ &=-(x+y+z)t+(xy+xz+yz)t^2-(xyz)t^3\\ &=-3t+(xy+xz+yz)t^2-(xyz)t^3 \end{align}

In the second equality we used the sum of the geometric series more or less like your friend said.

Comparing coefficients of the multiplication $FG$ with the result in the last line we get (the Newton's identities for the values in this problem)

\begin{align} 2(xy+xz+yz)&=9-5\\ 3xyz&=3(xy+xz+yz)-15+7\\ 0&=3(xyz)-5(xy+xz+yz)+21-(x^4+y^4+z^4) \end{align}

From these, substituting from top to bottom, we find the value of $x^4+y^4+z^4$. Notice that the method allows us to continue using the equalities between coefficients of higher degree in $t$ to compute $x^5+y^5+z^5$, $x^6+y^6+z^6$, and so on ...