Page:Squaring the circle a history of the problem (IA squaringcirclehi00hobsuoft).djvu/69

 The general value of $$\beta$$ consists of the sum of $$r$$ of the letters $$\alpha_1, \alpha_2, \ldots \alpha_s$$; and we consider those values of $$\beta$$ that correspond to a fixed value of $$r$$ to belong to one set. It is clear that a symmetrical function of those letters $$\beta$$ which belong to one and the same set is expressible as a symmetrical function of $$\alpha_1, \alpha_2, \ldots \alpha_s$$; therefore a symmetrical function of the products $$C\beta$$, where all the $$\beta$$'s belong to one and the same set, is in virtue of what has been established in (1) an integer. Applying the above lemma to all the $$n$$ numbers $$C\beta$$, we see that the symmetrical products formed by all the numbers $$C\beta$$ are integral, or zero. We have supposed those of the numbers $$\beta$$ which vanish to be suppressed and the corresponding exponentials to be absorbed in the integer $$A$$; whether this is done before or after the symmetrical functions of $$C\beta$$ are formed makes no difference, so that the above reasoning applies to the numbers $$C\beta$$ when those of them which vanish are removed.

(3) Let $$p$$ be a prime number greater than all the numbers $$A$$, $$n$$, $$C$$ $$|C^n \beta_1 \beta_2 \ldots \beta_n|$$; and let

We observe that $$\phi(x)$$ is of the form

where $$q_1, q_2, \ldots q_n$$ are integers. The function $$\phi(x)$$ may be expressed in the form

where $$c_{p-1} (p - 1)!, c_p p!, \ldots$$ are integral.

We see that $$\phi^{p-1}(0) = (-1)^{np} C^{p-1} q_n^p$$, which is an integer not divisible by $$p$$.

Also $$\phi^p(0)$$ is the value when $$x = 0$$ of

and is clearly an integer divisible by $$p$$. We see also that

are all multiples of $$p$$.

Further if $$m \leq n$$, $$\phi(\beta_m), \phi'(\beta_m), \ldots \phi^{(p-1)}(\beta_m)$$ all vanish, and