Elements of Algebra

CHAP. XV. OF ALGEBRA. 451 We shall begin by observing, that if this impossibility applies to the sum, it applies also to the difference, of two cubes. In fact, if it be impossible for $x^3 + y^3 = z^3$, it is also impossible for $z^3 - y^3 = x^3$. Now, $z^3 - y^3$ is the difference of two cubes; therefore, if the one be possible, the other is so likewise. This being laid down, it will be sufficient, if we demonstrate the impossibility either in the case of the sum, or difference; which demonstration requires the following chain of reasoning. 1. We may consider the numbers $x$ and $y$ as prime to each other; for if they had a common divisor, the cubes would also be divisible by the cube of that divisor. For example, let $x = ma$, and $y = mb$, we shall then have $x^3 + y^3 = m^3a^3 + m^3b^3$; now if this formula be a cube, $a^3 + b^3$ is a cube also. 2. Since, therefore, $x$ and $y$ have no common factor, these two numbers are either both odd, or the one is even and the other odd. In the first case, $z$ would be even, and in the other that number would be odd. Consequently, of these three numbers $x$, $y$, and $z$, there is always one which is even, and two that are odd; and it will therefore be sufficient for our demonstration to consider the case in which $x$ and $y$ are both odd: because we may prove the impossibility in question either for the sum, or for the difference; and the sum only happens to become the difference, when one of the roots is negative. 3. If therefore $x$ and $y$ are odd, it is evident that both their sum and their difference will be an even number. Therefore let $\frac{x+y}{2} = p$, and $\frac{x-y}{2} = q$, and we shall have $x = p + q$, and $y = p - q$; whence it follows, that one of the two numbers, $p$ and $q$, must be even, and the other odd. Now, we have, by adding $(p+q)^3 = x^3$, to $(p-q)^3 = y^3$, $x^3 + y^3 = 2p^3 + 6pq^2 = 2p(p^2 + 3q^2)$; so that it is required to prove that this product $2p(p^2 + 3q^2)$ cannot become a cube; and if the demonstration were applied to the difference, we should have $x^3 - y^3 = 6p^2q + 2q^3 = 2q(q^2 + 3p^2)$, a formula precisely the same as the former, if we substitute $p$ and $q$ for each other. Consequently, it is sufficient for our purpose to demonstrate the impossibility of the formula $2p(p^2 + 3q^2)$, since it will necessarily follow, that neither the sum nor the difference of two cubes can become a cube. 4. If therefore $2p(p^2 + 3q^2)$ were a cube, that cube would be even, and, consequently, divisible by 8: con- G G 2