Elements of Algebra

CHAP. VII. OF ALGEBRA. 353 than $\frac{2p}{q}$, or than $p$, let us suppose $n = p + q$, and we shall have, from the equation, $n = \frac{p + \sqrt{(3p^2 - 2)}}{2}$, $2p + 2q = p + \sqrt{(3p^2 - 2)}$, or $p + 2q = \sqrt{(3p^2 - 2)}$; then, by squaring, $p^2 + 4pq + 4q^2 = 3p^2 - 2$; so that $2p^2 = 4pq + 4q^2 + 2$, or $p^2 = 2pq + 2q^2 + 1$, and $p = q + \sqrt{(3q^2 + 1)}$. Now, this formula being similar to the one proposed, we may make $q = 0$, and shall thus obtain $p = 1$, and $n = 1$; whence $\sqrt{(3n^2 + 1)} = 2$. 100. Let $a = 5$, that we may have to make a square of the formula $5n^2 + 1$, the root of which is greater than $2n$. We shall therefore suppose $\sqrt{(5n^2 + 1)} = 2n + p$, or $5n^2 + 1 = 4n^2 + 4np + p^2$; whence we obtain $n^2 = 4np + p^2 - 1$, and $n = 2p + \sqrt{(5p^2 - 1)}$. Now, $\sqrt{(5p^2 - 1)} > 2p$; whence it follows that $n > 4p$; for which reason, we shall make $n = 4p + q$, which gives $2p + q = \sqrt{(5p^2 - 1)}$, or $4p^2 + 4pq + q^2 = 5p^2 - 1$, and $p^2 = 4pq + q^2 + 1$; so that $p = 2q + \sqrt{(5q^2 + 1)}$; and as $q = 0$ satisfies the terms of this equation, we shall have $p = 1$, and $n = 4$; therefore $\sqrt{(5n^2 + 1)} = 9$. 101. Let us now suppose $a = 6$, that we may have to consider the formula $6n^2 + 1$, whose root is likewise contained between $2n$ and $3n$. We shall, therefore, make $\sqrt{(6n^2 + 1)} = 2n + p$, and shall have $6n^2 + 1 = 4n^2 + 4np + p^2$, or $2n^2 = 4np + p^2 - 1$; and, thence, $n = p + \frac{\sqrt{(6p^2 - 2)}}{2}$, or $n = \frac{2p + \sqrt{(6p^2 - 2)}}{2}$; so that $n > 2p$. If, therefore, we make $n = 2p + q$, we shall have $4p + 2q = 2p + \sqrt{(6p^2 - 2)}$, or $2p + 2q = \sqrt{(6p^2 - 2)}$; the squares of which are $4p^2 + 8pq + 4q^2 = 6p^2 - 2$; so that $2p^2 = 8pq + 4q^2 + 2$, and $p^2 = 4pq + 2q^2 + 1$. Lastly, $p = 2q + \sqrt{(6q^2 + 1)}$. Now, this formula resembling the first, we have $q = 0$; wherefore $p = 1$, $n = 2$, and $\sqrt{(6n^2 + 1)} = 5$. 102. Let us proceed farther, and take $a = 7$, and $7n^2 + 1 = m^2$; here we see that $m > 2n$; let us therefore make $m = 2n + p$, and we shall have A A