Elements of Algebra

CHAP. XII. OF ALGEBRA. 403 $(p\sqrt{2} + q\sqrt{5})^3 = (2\sqrt{2} + \sqrt{5})^3 = 46\sqrt{2} + 29\sqrt{5}$, which is by no means the same as $4\sqrt{2} + \sqrt{5}$. But it must be remarked, that the formula $r^2 - 10s^2$ may become $1$, or $-1$, in an infinite number of cases; for example, if $r=3$, and $s=1$, or if $r=19$, and $s=6$: and this formula, multiplied by $2p^2 - 5q^2$, reproduces a number of this last form. Therefore, let $f^2 - 10g^2 = 1$; and, instead of supposing, as we have hitherto done, $2x^2 - 5y^2 = (2p^2 - 5q^2)^3$, we may suppose, in a more general manner, $2x^2 - 5y^2 = (f^2 - 10g^2) \times (2p^2 - 5q^2)^3$; so that, taking the factors, we shall have $x\sqrt{2} \pm y\sqrt{5} = (f \pm g\sqrt{10}) \times (p\sqrt{2} \pm q\sqrt{5})^3$. Now, $(p\sqrt{2} \pm q\sqrt{5})^3 = (2p^3 + 15pq^2)\sqrt{2} \pm (6p^2q + 5q^3)\sqrt{5}$; and if, in order to abridge, we write $A\sqrt{2} + B\sqrt{5}$ instead of this quantity, and multiply by $f + g\sqrt{10}$, we shall have $Af\sqrt{2} + Bf\sqrt{5} + 2Ag\sqrt{5} + 5Bg\sqrt{2}$ to make equal to $x\sqrt{2} + y\sqrt{5}$; whence results $x = Af + 5Bg$, and $y = Bf + 2Ag$. Now, since we must have $y = \pm 1$, it is not absolutely necessary that $6p^2q + 5q^3 = 1$; on the contrary, it is sufficient that the formula $Bf + 2Ag$, that is to say, that $f(6p^2q + 5q^3) + 2g(2p^3 + 15pq^2)$ becomes $= \pm 1$; so that $f$ and $g$ may have several values. For example, let $f=3$, and $g=1$, the formula $18p^2q + 15q^3 + 4p^3 + 30pq^2$ must become $\pm 1$; that is, $4p^3 + 18p^2q + 30pq^2 + 15q^3 = \pm 1$. 197. The difficulty, however, of determining all the possible cases of this kind, exists only in the formula $ax^2 + cy^2$, when the number $c$ is negative; and the reason is, that this formula, namely, $x^2 - acy^2$, which depends on it, may then become $1$; which never happens when $c$ is a positive number, because, $x^2 + cy^2$, or $x^2 + acy^2$, always gives greater numbers, the greater the values we assign to $x$ and $y$. For which reason, the method we have explained cannot be successfully employed, except in those cases in which the two numbers $a$ and $c$ have positive values. 198. Let us now proceed to the fourth degree. Here we shall begin by observing, that if the formula $ax^2 + cy^2$ is to be changed into a biquadrate, we must have $a=1$; for it would not be possible even to transform the formula into a square (Art. 181); and, if this were possible, we might also give it the form $t^2 + acu^2$; for which reason we shall extend the question only to this last formula, which may be reduced to the former, $x^2 + cy^2$, by supposing $a=1$. This DD2