Elements of Algebra

the values of $x$ and $y$, they will likewise be found to be double; that is to say, we shall have $x = pr + cqs$, and $y = qr + ps$; we shall also have $x = pr - cqs$, and $y = ps - qr$. If we chose to make trial, and see whether we obtain from these values the product already found, we should have, by trying the first, $x^2 = p^2r^2 + 2cpqrs + c^2q^2s^2$, and $y^2 = p^2s^2 + 2pqrs + q^2r^2$, or $cy^2 = cp^2s^2 + 2cpqrs + cq^2r^2$; so that $x^2 - cy^2 = p^2r^2 - cp^2s^2 + c^2q^2s^2 - cq^2r^2$, which is just the product already found, $(p^2 - cq^2) \times (r^2 - cs^2)$. 177. Hitherto we have considered the first term as without a coefficient; but we shall now suppose that term to be multiplied also by another letter, and shall seek what factors the formula $ax^2 + cy^2$ may contain. Here it is evident that our formula is equal to the product $(x\sqrt{a} + y\sqrt{-c}) \times (x\sqrt{a} - y\sqrt{-c})$, and, consequently, that it is required to give factors also to these two factors. Now, in this a difficulty occurs; for if, according to the second method, we make $x\sqrt{a} + y\sqrt{-c} = (p\sqrt{a} + q\sqrt{-c}) \times (r\sqrt{a} + s\sqrt{-c}) = apr - cqs + ps\sqrt{-ac} + qr\sqrt{-ac}$, and $x\sqrt{a} - y\sqrt{-c} = (p\sqrt{a} - q\sqrt{-c}) \times (r\sqrt{a} - s\sqrt{-c}) = apr - cqs - ps\sqrt{-ac} - qr\sqrt{-ac}$, we shall have $2x\sqrt{a} = 2apr - 2cqs$, and $2y\sqrt{-c} = 2ps\sqrt{-ac} + 2qr\sqrt{-ac}$; that is to say, we have found both for $x$ and for $y$ irrational values, which cannot here be admitted. 178. But this difficulty may be removed thus: let us make $x\sqrt{a} + y\sqrt{-c} = (p\sqrt{a} + q\sqrt{-c}) \times (r + s\sqrt{-ac}) = pr\sqrt{a} - cqs\sqrt{a} + qr\sqrt{-c} + aps\sqrt{-c}$, and $x\sqrt{a} - y\sqrt{-c} = (p\sqrt{a} - q\sqrt{-c}) \times (r - s\sqrt{-ac}) = pr\sqrt{a} - cqs\sqrt{a} - qr\sqrt{-c} - aps\sqrt{-c}$. This supposition will give the following values for $x$ and $y$; namely, $x = pr - cqs$, and $y = qr + aps$; and our formula, $ax^2 + cy^2$, will have the factors $(ap^2 + cq^2) \times (r^2 + acs^2)$, one of which only is of the same form with the formula, the other being different. 179. There is still, however, a great affinity between these two formulæ, or factors; since all the numbers contained in the first, if multiplied by a number contained in the second, revert again to the first. We have already seen, that two numbers of the second form $x^2 + acy^2$, which