UTM Ideals Varieties and Algorithm - Chapter 1 Section 3 Exercise 7

Problem:

Solution:

The generalization is pretty obvious, we claim that

$x_j = \frac{2u_j}{(\sum\limits_{i = 1}^{n-1}{u_i^2}) + 1}$ for $j = 1 \cdots n - 1$.

$x_n = \frac{(\sum\limits_{i = 1}^{n-1}{u_i^2}) - 1}{(\sum\limits_{i = 1}^{n-1}{u_i^2}) + 1}$ for $j = n$.

All we need to do is to prove that it is correct by substitute that into the equation $x_1^2 + \cdots + x_n^2$ and show that it is identically 1.

$\begin{eqnarray*} & & x_1^2 + \cdots + x_n^2 \\ &=& (\sum\limits_{j = 1}^{n-1}{x_j^2}) + x_n^2 \\ &=& \left(\sum\limits_{j = 1}^{n-1}{\left(\frac{2u_j}{(\sum\limits_{i = 1}^{n-1}{u_i^2}) + 1}\right)^2}\right) + \left(\frac{(\sum\limits_{i = 1}^{n-1}{u_i^2}) - 1}{(\sum\limits_{i = 1}^{n-1}{u_i^2}) + 1}\right)^2 \\ \end{eqnarray*}$

To make our notation simpler, let $S = \sum\limits_{i = 1}^{n-1}{u_i^2}$, so we can simplify to

$\begin{eqnarray*} &=& \left(\sum\limits_{j = 1}^{n-1}{\left(\frac{2u_j}{S + 1}\right)^2}\right) + \left(\frac{S - 1}{S + 1}\right)^2 \\ &=& \frac{1}{(S+1)^2}(\sum\limits_{j = 1}^{n-1}{(2u_j)^2} + (S - 1)^2) \\ &=& \frac{1}{(S+1)^2}(4S + (S - 1)^2) \\ &=& \frac{1}{(S+1)^2}(4S + S^2 - 2S + 1) \\ &=& \frac{1}{(S+1)^2}(S^2 + 2S + 1) \\ &=& \frac{1}{(S+1)^2}(S + 1)^2 \\ &=& 1 \end{eqnarray*}$