Elements of Algebra

440 ELEMENTS PART II. although neither $5 + 3$, nor $5 - 3$, are squares. Let us, therefore, resolve the question independently of this consideration, and remark, in the first place, that we may take $1$ for one of the squares sought: the reason for which is, that if the formulæ $x^2 - y^2$, $x^2 - z^2$, and $y^2 - z^2$, are squares, they will continue so, though divided by $z^2$; consequently, we may suppose that the question is to transform $(\frac{x^2}{z^2} - \frac{y^2}{z^2})$, $(\frac{x^2}{z^2} - 1)$, and $(\frac{y^2}{z^2} - 1)$ into squares, and it then refers only to the two fractions $\frac{x}{z}$, and $\frac{y}{z}$. If we now suppose $\frac{x}{z} = \frac{p^2 + 1}{p^2 - 1}$, and $\frac{y}{z} = \frac{q^2 + 1}{q^2 - 1}$, the last two conditions will be satisfied; for we shall then have $\frac{x^2}{z^2} - 1 = \frac{4p^2}{(p^2 - 1)^2}$, and $\frac{y^2}{z^2} - 1 = \frac{4q^2}{(q^2 - 1)^2}$. It only remains, therefore, to consider the first formula $\frac{x^2}{z^2} - \frac{y^2}{z^2} = \frac{(p^2+1)^2}{(p^2-1)^2} - \frac{(q^2+1)^2}{(q^2-1)^2} =$ $(\frac{p^2+1}{p^2-1} + \frac{q^2+1}{q^2-1}) \times (\frac{p^2+1}{p^2-1} - \frac{q^2+1}{q^2-1})$. Now, the first factor here is $\frac{2(p^2q^2 - 1)}{(p^2-1) \times (q^2-1)}$; the second is $\frac{2(q^2 - p^2)}{(p^2-1) \times (q^2-1)}$, and the product of these two factors is $= \frac{4(p^2q^2-1) \times (q^2-p^2)}{(p^2-1)^2 \times (q^2-1)^2}$. It is evident that the denominator of this product is already a square, and that the numerator contains the square $4$; therefore it is only required to transform into a square the formula $(p^2q^2 - 1) \times (q^2 - p^2)$, or $(p^2q^2 - 1) \times (\frac{q^2}{p^2} - 1)$; and this is done by making $pq = \frac{f^2 + g^2}{2fg}$, and $\frac{q}{p} = \frac{h^2 + k^2}{2hk}$, because then each factor separately becomes a square. We may also be convinced of this, by remarking that $pq \times \frac{q}{p} = q^2 = \frac{f^2+g^2}{2fg} \times \frac{h^2+k^2}{2hk}$; and, consequently, the product of these two fractions must be a square; as it must also be when multiplied by