Page:Squaring the circle a history of the problem (IA squaringcirclehi00hobsuoft).djvu/68

are positive or negative integers (including zero); thus one of the numbers $\alpha _{1},\ldots \alpha _{s}$ is $i\pi$.

From Euler's equation $e^{i\pi }+1=0$, we see that the relation

$(1+e^{\alpha _{1}})(1+e^{\alpha _{2}})\ldots (1+e^{\alpha _{n}})=0$

must hold, since one of the factors vanishes. If we multiply out the factors in this equation, it clearly takes the form

$A+e^{\beta _{1}}+e^{\beta _{2}}+\ldots +e^{\beta _{n}}=0$

where A is some positive integer (≥ 1), being made up of 1 together with those terms, if any, which are of the form $e^{\alpha _{p}+\alpha _{q}+\ldots }$, where

$\alpha _{p}+\alpha _{q}+\ldots =0$.

(2) A symmetrical function consisting of the sum of the products taken in every possible way, of a fixed number of the numbers $C\alpha _{1},C\alpha _{2},\ldots C\alpha _{s}$, is an integer. It will be proved that the symmetrical functions of $C\beta _{1},C\beta _{2},\ldots C\beta _{n}$ have the same property. In order to prove this we have need of the following lemma:

A symmetrical function consisting of the sums of the products taken $p$ together of $\alpha +\beta +\gamma +\ldots$ letters

$x_{1},x_{2},\ldots x_{\alpha }$; $y_{1},y_{2},\ldots y_{\beta }$; $z_{1},z_{2},\ldots z_{\gamma }$; &c.,

belonging to any number of separate sets, can be expressed in terms of symmetrical functions of the letters in the separate sets.

It will be sufficient to prove this in the case in which there are only two sets of letters, the extension to the general case being then obvious.

Denote by $\textstyle \sum \limits _{p}P(x,y)$ the sum of the products which we require to express; and denote by $\textstyle \sum \limits _{r}P(x)$ the sum of the products of $r$ dimensions of the letters $x_{1},x_{2},\ldots x_{\alpha }$ only. In case $p\geq \alpha$, we see that

$\textstyle \sum \limits _{p}P(x,y)=\sum \limits _{p}P(x)+\sum \limits _{1}P(y)\sum \limits _{p-1}P(x)+\sum \limits _{2}P(y)\sum \limits _{p-2}P(x)+\ldots$;

in case $p>\alpha$, we see that

$\textstyle \sum \limits _{p}P(x,y)=\sum \limits _{\alpha }P(x)\sum \limits _{p-\alpha }P(y)+\sum \limits _{\alpha -1}P(x)\sum \limits _{p-\alpha +1}P(y)+\sum \limits _{\alpha -2}P(x)\sum \limits _{p-\alpha +2}P(y)+\ldots$;

and the terms on the right-hand side involve in each case only symmetrical functions of the letters of the two separate sets; thus the lemma is established.

To apply this lemma, we observe that the numbers $\beta$ fall into separate sets, according to the way they are formed from the letters $\alpha$.