Elements of Algebra

CHAP. I. OF ALGEBRA. 311 question) for the two others, we must, in consequence of that result, have $N = 209u + 157$, instead of which we shall write $N = 209q + 157$; so that $29p + 10 = 209q + 157$, or $29p = 209q + 147$; whence we have the following Table; $209 = 7 \times 29 + 6;$ $29 = 4 \times 6 + 5;$ $6 = 1 \times 5 + 1;$ $5 = 5 \times 1 + 0;$ wherefore $\begin{cases} p = 7q + r, \\ q = 4r + s, \\ r = s + t, \\ s = 5t - 147. \end{cases}$ And, if we now retrace these steps, we have $s = 5t - 147,$ $r = s + t = 6t - 147,$ $q = 4r + s = 29t - 735,$ $p = 7q + r = 209t - 5292.$ So that $N = 6061t - 153458$*; and the least number is found by making $t = 26$, which supposition gives $N = 4128$. 22. It is necessary, however, to observe, in order that an equation of the form $bp = aq + n$ may be resolvible, that the two numbers $a$ and $b$ must have no common divisor; for, otherwise, the question would be impossible, unless the number $n$ had the same common divisor also. If it were required, for example, to have $9p = 15q + 2$; since 9 and 15 have a common divisor 3, and which is not a divisor of 2, it is impossible to resolve the question, because $9p - 15q$ being always divisible by 3, can never become $= 2$. But if in this example $n = 3$, or $n = 6$, &c. the question would be possible: for it would be sufficient first to divide by 3; since we should obtain $3p = 5q + 1$, an equation easily resolvible by the rule already given. It is evident, therefore, that the numbers $a$, $b$, ought to have no common divisor, and that our rule cannot apply in any other case. 23. To prove this still more satisfactorily, we shall consider the equation $9p = 15q + 2$ according to the usual method. Here we find $p = \frac{15q+2}{9} = q + \frac{6q+2}{9} = q + r$; so that $9r = 6q + 2$, or $6q = 9r - 2$; or $q = \frac{9r-2}{6} = r + \frac{3r-2}{6} = r + s$; so that $3r - 2 = 6s$, * That is, $-5292 \times 29 = -153468$; to which if the remainder $+10$ required by the question be added, the sum is $-153458$.