Elements of Algebra

gives $2a^3 = y^3 - x^3$; it is likewise impossible to find two cubes having their difference equal to the double of another cube; and the same impossibility extends to the sum of two cubes, as is evident from the following demonstration. 247. *Theorem*. Neither the sum nor the difference of two cubes can become equal to the double of another cube; or, in other words, the formula $x^3 \pm y^3 = 2z^3$ is always impossible, except in the evident case of $y=x$. We may here also consider $x$ and $y$ as prime to each other; for if these numbers had a common divisor, it would be necessary for $z$ to have the same divisor; and, consequently, for the whole equation to be divisible by the cube of that divisor. This being laid down, as $x^3 \pm y^3$ must be an even number, the numbers $x$ and $y$ must both be odd, in consequence of which both their sum and their difference must be even. Making, therefore, $\frac{x+y}{2} = p$, and $\frac{x-y}{2} = q$, we shall have $x=p+q$ and $y=p-q$; and of the two numbers $p$ and $q$, the one must be even and the other odd. Now, from this, we obtain $x^3+y^3 = 2p^3 + 6pq^2 = 2p(p^2 + 3q^2)$, and $x^3-y^3 = 6p^2q + 2q^3 = 2q(3p^2 + q^2)$, which are two formulæ perfectly similar. It will therefore be sufficient to prove that the formula $2p(p^2+3q^2)$ cannot become the double of a cube, or that $p(p^2+3q^2)$ cannot become a cube: which may be demonstrated in the following manner. 1. Two different cases again present themselves to our consideration: the one, in which the two factors $p$, and $p^2+3q^2$, have no common divisor, and must separately be a cube; the other in which these factors have a common divisor, which divisor, however, as we have seen (Art. 243), can be no other than 3. 2. *Case* 1. Supposing, therefore, that $p$ is not divisible by 3, and that thus the two factors are prime to each other, we shall first reduce $p^2+3q^2$ to a cube by making $p=t(t^2-9u^2)$, and $q=3u(t^2-9u^2)$; by which means it will only be farther necessary for $p$ to become a cube. Now, $t$ not being divisible by 3, since otherwise $p$ would also be divisible by 3, the two factors $t$, and $t^2-9u^2$, are prime to one another, and, consequently, each must separately be a cube. 3. But the last factor has also two factors, namely $t+3u$, and $t-3u$, which are prime to each other, first because $t$ is not divisible by 3, and, in the second place, because one of