Elements of Algebra

498 ADDITIONS. CHAP. II. and are $p$ in number, it is evident that $0$ must be among those remainders, and, consequently, that there is one of the numbers $q \pm 1$, $2q \pm 1$, $3q \pm 1$, &c. $pq \pm 1$, which is divisible by $p$. Now, it is evident that this cannot be the last; so that there is certainly a value of $r$ less than $p$, which will make $rq \pm 1$ divisible by $p$; and it is evident, at the same time, that the quotient will be less than $q$; therefore there will always be an integer and positive value of $r$ less than $p$, and another similar value of $s$, and less than $q$, which will satisfy the equation $s = \frac{qr \pm 1}{p}$, or $ps - qr = \pm 1$. 24. The question is therefore now reduced to this; to find four positive whole numbers, $p, q, r, s$, the last two of which may be less than the first two; that is, $r < p$, and $s < q$, and such, that $ps - qr = \pm 1$; farther, that the quantities $p-aq$, and $r-as$, may have different signs, and, at the same time, that $r-as$ may be a quantity greater than $p-aq$, abstracting from the signs. In order to simplify, let us denote $r$ by $p'$, and $s$ by $q'$, so that we have $pq' - qp' = \pm 1$; and as $q > q'$ (hyp.), let $\mu$ be the quotient that would be produced by the division of $q$ by $q'$, and let the remainder be $q''$, which will consequently be $< q'$; also, let $\mu'$ be the quotient of the division of $q'$ by $q''$, and $q'''$ the remainder, which will be $< q''$; in like manner, let $\mu''$ be the quotient of the division of $q''$ by $q'''$, and $q^{iv}$ the remainder $< q'''$, and so on, till there is no remainder; in this way, we shall have $q = \mu q' + q''$ $q' = \mu' q'' + q'''$ $q'' = \mu'' q''' + q^{iv}$ $q''' = \mu''' q^{iv} + q^v$, &c. where the numbers $\mu, \mu', \mu''$, &c. will all be integer and positive, and the numbers $q', q'', q'''$, &c. will also be integer and positive, and will form a series decreasing to nothing. In like manner, let us suppose $p = \mu p' + p''$ $p' = \mu' p'' + p'''$ $p'' = \mu'' p''' + p^{iv}$ $p''' = \mu''' p^{iv} + p^v$, &c. And as the numbers $p$ and $p'$ are considered here as given, as well as the numbers $\mu, \mu', \mu''$, &c. we may determine from these equations the numbers $p'', p''', p^{iv}$, &c. which will evidently be all integer.