Elements of Algebra

CHAP. III. ADDITIONS. 533 Then, with the quotients 1, 2, 3, \ldots we may form the fractions, 1, 2, 3, 2, 2. $\frac{1}{1}$, $\frac{3}{2}$, $\frac{10}{7}$, $\frac{23}{16}$, $\frac{56}{39}$. and the last fraction but one, $\frac{23}{16}$, will be that which we have expressed in general by $\frac{p}{q}$; so that we shall have $p=23$, $q=16$; and, as this fraction is the fourth, and consequently, of an even rank, we must take the upper sign; so that we shall have, in general, $x = 23 \times 11 + 56m$, and $y = 16 \times 11 + 39m$; $m$ being any integer number whatever, positive, or negative. 45. *Scholium*. We owe the first solution of this problem to M. Bachet de Meziriac, who gave it in the second edition of his Mathematical Recreations, entitled *Problemes plaisans et delectables*, \ldots The first edition of this work appeared in 1612; but the solution in question is there only announced, and is only found complete in the edition of 1624. The method of M. Bachet is very direct and ingenious, and cannot be rendered more elegant, or more general. I seize with pleasure the present opportunity of doing justice to this learned author, having observed that the mathematicians, who have since resolved the same problem, have never taken any notice of his labours. The method of M. Bachet may be explained in a few words. After having shewn how the solution of equations of the form $ax - by = c$, ($a$ and $b$ being prime to each other), may be reduced to that of $ax - by = \pm 1$, he applies to the resolution of this last equation; and, for this purpose, prescribes the same operation with regard to the numbers $a$ and $b$, as if we wished to find their greatest common divisor, (and this is what we have just done); then calling $c, d, e, f, \ldots$ the remainders arising from the different divisions, and supposing, for example, that $f$ is the last remainder, which will necessarily be equal to unity (because $a$ and $b$ are prime to one another, by hypothesis), he makes, when the number of remainders is even, as in the present case, $e \mp 1 = \epsilon$, $\frac{\epsilon d \pm 1}{e} = \delta$, $\frac{\delta c \mp 1}{d} = \gamma$, $\frac{\gamma b \pm 1}{c} = \beta$, $\frac{\beta a \mp 1}{b} = \alpha$;