Elements of Algebra

352 ELEMENTS PART II. and this, Pell, an English writer, has taught us to find by an ingenious method, which we shall here explain. 98. This method is not such as may be employed generally, for any number $a$ whatever; it is applicable only to each particular case. We shall therefore begin with the easiest cases, and shall first seek such a value of $n$, that $2n^2 + 1$ may be a square, or that $\sqrt{(2n^2 + 1)}$ may become rational. We immediately see that this square root becomes greater than $n$, and less than $2n$. If, therefore, we express this root by $n + p$, it is obvious that $p$ must be less than $n$; and we shall have $\sqrt{(2n^2 + 1)} = n + p$; then, by squaring, $2n^2 + 1 = n^2 + 2np + p^2$; or $n^2 = 2pn + p^2 - 1$; therefore, by compleating the square, \ldots $n^2 = 2pn + p^2 - 1$, and $n = p + \sqrt{(2p^2 - 1)}$. The whole is reduced, therefore, to the condition of $2p^2 - 1$ being a square; now, this is the case if $p = 1$, which gives $n = 2$, and $\sqrt{(2n^2 + 1)} = 3$. If this case had not been immediately obvious, we should have gone farther; and since $\sqrt{(2p^2 - 1)} \succ p$*, and, consequently, $n \succ 2p$, we should have made $n = 2p + q$; and should thus have had $2p + q = p + \sqrt{(2p^2 - 1)}$, or $p + q = \sqrt{(2p^2 - 1)}$, and, squaring, $p^2 + 2pq + q^2 = 2p^2 - 1$, whence $p^2 = 2pq + q^2 + 1$, which would have given $p = q + \sqrt{(2q^2 + 1)}$; so that it would have been necessary to have $2q^2 + 1$ a square; and as this is the case, if we make $q = 0$, we shall have $p = 1$, and $n = 2$, as before. This example is sufficient to give an idea of the method; but it will be rendered more clear and distinct from what follows. 99. Let $a = 3$, that is to say, let it be required to transform the formula $3n^2 + 1$ into a square. Here we shall make $\sqrt{(3n^2 + 1)} = n + p$, which gives $3n^2 + 1 = n^2 + 2np + p^2$, and $2n^2 = 2np + p^2 - 1$; whence we obtain $n = \frac{p + \sqrt{(3p^2 - 2)}}{2}$. Now, since $\sqrt{(3p^2 - 2)}$ exceeds $p$, and, consequently, $n$ is greater * This sign, $\succ$, placed between two quantities, signifies that the former is greater than the latter; and when the angular point is turned the contrary way, as $\prec$, it signifies that the former is less than the latter.