Elements of Algebra

CHAP. VII. OF ALGEBRA. 355 ready similar to the one proposed, we may make $p = 0$, which gives $n = 1$, and $m = 3$. 104. We may proceed, in the same manner, for every other number, $a$, provided it be positive and not a square, and we shall always be led, at last, to a radical quantity, such as $\sqrt{(at^2 + 1)}$, similar to the first, or given formula, and then we have only to suppose $t = 0$; for the irrationality will disappear, and by tracing back the steps, we shall necessarily find such a value of $n$, as will make $an^2 + 1$ a square. Sometimes we quickly obtain our end; but, frequently also, we are obliged to go through a great number of operations. This depends on the nature of the number $a$; but we have no principles, by which we can foresee the number of operations that it will be necessary to perform. The process is not very long for numbers below 13, but when $a = 13$, the calculation becomes much more prolix; and, for this reason, it will be proper here to resolve that case. 105. Let therefore $a = 13$, and let it be required to find $13n^2 + 1 = m^2$. Here, as $m^2 > 9n^2$, and, consequently, $m > 3n$, let us suppose $m = 3n + p$; we shall then have $13n^2 + 1 = 9n^2 + 6np + p^2$, or $4n^2 = 6np + p^2 - 1$, and $n = \frac{3p + \sqrt{(13p^2 - 4)}}{4}$, which shews that $n > \frac{6}{4}p$, and therefore much greater than $p$. If, therefore, we make $n = p + q$, we shall have $p + 4q = \sqrt{(13p^2 - 4)}$; and, taking the squares, $13p^2 - 4 = p^2 + 8pq + 16q^2$; so that $12p^2 = 8pq + 16q^2 + 4$, or $3p^2 = 2pq + 4q^2 + 1$, and $p = \frac{q + \sqrt{(13q^2 + 3)}}{3}$. Here, $p > \frac{q + 3q}{3}$, or $p > q$; we shall proceed, therefore, by making $p = q + r$, and shall thus obtain $2q + 3r = \sqrt{(13q^2 + 3)}$; then $13q^2 + 3 = 4q^2 + 12qr + 9r^2$, or $9q^2 = 12qr + 9r^2 - 3$, or $3q^2 = 4qr + 3r^2 - 1$; which gives $q = \frac{2r + \sqrt{(13r^2 - 3)}}{3}$. Again, since $q > \frac{2r + 3r}{3}$, or $q > r$, we shall make $q = r + s$, and we shall thus have $r + 3s = \sqrt{(13r^2 - 3)}$; or $13r^2 - 3 = r^2 + 6rs + 9s^2$, or $12r^2 = 6rs + 9s^2 + 3$, or $4r^2 = 2rs + 3s^2 + 1$; whence we obtain AA 2