Elements of Algebra

324 ELEMENTS PART II. We have $x = \frac{3}{4}, \frac{4}{3}, \frac{5}{12}, \frac{15}{8}, \frac{7}{24}, \frac{12}{35}, \frac{21}{20}, \frac{8}{15}, \frac{6}{40}, \ldots$ 44. We have, therefore, in general, [Art. 42, 43.] $1 + \frac{(n^2 - m^2)^2}{(2mn)^2} = \frac{(n^2 + m^2)^2}{(2mn)^2}$; and, if we multiply this equation by $(2mn)^2$, we find $(2mn)^2 + (n^2 - m^2)^2 = (n^2 + m^2)^2$; so that we know, in a general manner, two squares, whose sum gives a new square. This remark will lead to the solution of the following question: To find two square numbers, whose sum is likewise a square number. We must have $p^2 + q^2 = r^2$; we have therefore only to make $p = 2mn$, and $q = n^2 - m^2$, then we shall have $r = n^2 + m^2$. Farther, as $(n^2 + m^2)^2 - (2mn)^2 = (n^2 - m^2)^2$, we may also resolve the following question: To find two squares, whose difference may also be a square number. Here, since $p^2 - q^2 = r^2$, we have only to suppose $p = n^2 + m^2$, and $q = 2mn$, and we obtain $r = n^2 - m^2$. We might also make $p = n^2 + m^2$, and $q = n^2 - m^2$, from which we should find $r = 2mn$. 45. We spoke of two methods of giving the form of a square to the formula $1 + x^2$. The other is as follows: If we suppose $\sqrt{(1 + x^2)} = 1 + \frac{mx}{n}$, we shall have $1 + x^2 = 1 + \frac{2mx}{n} + \frac{m^2x^2}{n^2}$; subtracting $1$ from both sides, $x^2 = \frac{2mx}{n} + \frac{m^2x^2}{n^2}$. This equation being divided by $x$, we have $x = \frac{2m}{n} + \frac{m^2x}{n^2}$, or $n^2x = 2mn + m^2x$, whence $x = \frac{2mn}{n^2 - m^2}$. Having found this value of $x$, we have * Thus, if $n = 3$, and $m = 2$, we have, by the last equation, $\sqrt{(1 + x^2)} = \frac{3^2 + 2^2}{2(3 \times 2)} = \frac{13}{12}$; or $1 + x^2 = \frac{13^2}{12^2}$, and $x^2 = \frac{13^2}{12^2} - 1$. Then $x = \sqrt{(\frac{13^2}{12^2} - 1)}$; that is, $x = \sqrt{\frac{25}{144}} = \frac{5}{12}$, as above.