Elements of Algebra

CHAP. XV. OF ALGEBRA. 453 found $p = t(t^2 - 9u^2) = t(t + 3u) \times (t - 3u)$, it is also required that $\frac{p}{4}$, and consequently $2p$, be a cube; or, which comes to the same, that the formula $2t(t + 3u) \times (t - 3u)$ be a cube. But here it must be observed that $t$ is an even number, and not divisible by 3; since otherwise $p$ would be divisible by 3, which we have expressly supposed not to be the case: so that the three factors, $2t$, $t + 3u$, and $t - 3u$, are prime to each other; and each of them must separately be a cube. If, therefore, we make $t + 3u = f^3$, and $t - 3u = g^3$, we shall have $2t = f^3 + g^3$. So that, if $2t$ is a cube, we shall have two cubes $f^3$, and $g^3$, whose sum would be a cube, and which would evidently be much less than the cubes $x^3$ and $y^3$ assumed at first; for as we first made $x = p + q$, and $y = p - q$, and have now determined $p$ and $q$ by the letters $t$ and $u$, the numbers $x$ and $y$ must necessarily be much greater than $t$ and $u$. 9. If, therefore, there could be found in great numbers two such cubes as we require, we should also be able to assign in less numbers two cubes whose sum would make a cube, and in the same manner we should be led to cubes always less. Now, as it is very certain that there are no such cubes among small numbers, it follows that there are not any among the greater numbers. This conclusion is confirmed by that which the second case furnishes, and which will be seen to be the same. 10. Case 2. Let us now suppose, that $p$ is divisible by 3, and that $q$ is not so, and let us make $p = 3r$; our formula will then become $\frac{3r}{4} \times (9r^2 + 3q^2)$, or $2r(3r^2 + q^2)$; and these two factors are prime to each other, since $3r^2 + q^2$ is neither divisible by 2 nor by 3, and $r$ must be even as well as $p$; therefore each of these two factors must separately be a cube. 11. Now, by transforming the second factor $3r^2 + q^2$, or $q^2 + 3r^2$, we find, in the same manner as before, $q = t(t^2 - 9u^2)$, and $r = 3u(t^2 - u^2)$; and it must be observed, that since $q$ was odd, $t$ must be here likewise an odd number, and $u$ must be even. 12. But $\frac{9r}{4}$ must also be a cube; or multiplying by the cube $\frac{8}{27}$, we must have $\frac{2r}{3}$, or