Elements of Algebra

400 ELEMENTS PART II. 189. Let us also consider the formula $x^2 + 3y^2$, for the purpose of making it equal to a cube. As we have, in this case, $a = 1$, and $c = 3$, we find $x = p^3 - 9pq^2$, and $y = 3p^2q - 3q^3$, whence $x^2 + 3y^2 = (p^2 + 3q^2)^3$. This formula occurs very frequently; for which reason we shall here give a Table of the easiest cases. $p$ | $q$ | $x$ | $y$ | $x^2 + 3y^2$ --|---|---|---|--- $1$ | $1$ | $8$ | $0$ | $64 = 4^3$ $2$ | $1$ | $10$ | $9$ | $343 = 7^3$ $1$ | $2$ | $35$ | $18$ | $2197 = 13^3$ $3$ | $1$ | $0$ | $24$ | $1728 = 12^3$ $1$ | $3$ | $80$ | $72$ | $21952 = 28^3$ $3$ | $2$ | $81$ | $30$ | $9261 = 21^3$ $2$ | $3$ | $154$ | $45$ | $29791 = 31^3$ 190. If the question were not restricted to the condition, that the numbers $x$ and $y$ must have no common divisor, it would not be attended with any difficulty; for if $ax^2 + cy^2$ were required to be a cube, we should only have to make $x = tz$, and $y = uz$, and the formula would become $at^2z^2 + cu^2z^2$; which we might make equal to the cube $v^3z^3$, and should immediately find $z = v^3(at^2 + cu^2)$. Consequently, the values sought of $x$ and $y$ would be $x = tv^3(at^2 + cu^2)$, and $y = uv^3(at^2 + cu^2)$, which, beside the cube $v^3$, have also the quantity $at^2 + cu^2$ for a common divisor; so that this solution immediately gives $ax^2 + cy^2 = v^6(at^2+cu^2)^2 \times (at^2+cu^2) = v^6(at^2+cu^2)^3$, which is evidently the cube of $v^2(at^2 + cu^2)$. 191. This last method, which we have made use of, is so much the more remarkable, as we are brought to solutions, which absolutely required numbers rational and integer, by means of irrational, and even imaginary quantities; and, what is still more worthy of attention, our method cannot be applied to those cases, in which the irrationality vanishes. For example, when the formula $x^2 + cy^2$ must become a cube, we can only infer from it, that its two irrational factors, $x + y\sqrt{-c}$, and $x - y\sqrt{-c}$, must likewise be cubes; and since $x$ and $y$ have no common divisor, these factors cannot have any. But if the radicals were to disappear, as in the case of $c = -1$, this principle would no