Elements of Algebra

CHAP. VI. OF ALGEBRA. 343 first, divided by $q$, gives $y + g = \frac{apx + apf}{q}$; and the second, divided by $p$, gives $y - g = \frac{qx - qf}{p}$. Subtracting this from the former, $2g = \frac{(ap^2 - q^2)x + (ap^2 + q^2)f}{pq}$, or $2gpq = (ap^2 - q^2)x + (ap^2 + q^2)f$; therefore $x = \frac{2gpq}{ap^2 - q^2} - \frac{(ap^2 + q^2)f}{ap^2 - q^2}$, from which, (by substituting this value of $x$, in the equation, $y - g = \frac{qx - qf}{p}$), we obtain $y = g + \frac{2gq^2}{ap^2 - q^2} - \frac{(ap^2 + q^2)fq}{(ap^2 - q^2)p} - \frac{qf}{p}$. In this latter value, as the first two terms, both containing the letter $g$, may be put into the form $\frac{g(ap^2 + q^2)}{ap^2 - q^2}$*, and as the other two, containing the letter $f$, may be expressed by $\frac{2afpq}{ap^2 - q^2}$, all the terms will be reduced to the same denomination, and we shall have $y = \frac{g(ap^2 + q^2) - 2afpq}{ap^2 - q^2}$. 82. This operation seems not, at first, to answer our purpose; since having to find integer values of $x$ and $y$, we are brought to fractional results; and it would be required to solve this new question,—What numbers are we to substitute for $p$ and $q$, in order that the fraction may disappear? A question apparently still more difficult than our original one: but here we may employ a particular artifice, which will readily bring us to our object, and which is as follows: As every thing must be expressed in integer numbers, let us make $\frac{ap^2 + q^2}{ap^2 - q^2} = m$, and $\frac{2pq}{ap^2 - q^2} = n$, in order that we may have $x = ng - mf$, and $y = mg - naf$. Now, we cannot here assume $m$ and $n$ at pleasure, since these letters must be such as will answer to what has been already determined: therefore, for this purpose, let us consider their squares, and we shall find * For $g = \frac{g(ap^2 - q^2)}{ap^2 - q^2} = \frac{gap^2 - gq^2}{ap^2 - q^2}$; and $\frac{2gq^2}{ap^2 - q^2} + \frac{gap^2 - gq^2}{ap^2 - q^2} = \frac{2gq^2 + gap^2 - gq^2}{ap^2 - q^2} = \frac{g(ap^2 + q^2)}{ap^2 - q^2}$.