Elements of Algebra

CHAP. XV. OF ALGEBRA. 457 the numbers $t$ or $u$ is even, and the other odd; for if these numbers were both odd, not only $p$, but also $q$, must be odd, which cannot be: therefore, each of these two factors, $t+3u$, and $t-3u$, must separately be a cube. 4. Therefore let $t+3u=f^3$, and $t-3u=g^3$, and we shall then have $2t=f^3+g^3$. Now, $t$ must be a cube, which we shall denote by $h^3$; by which means we must have $f^3+g^3=2h^3$; consequently, we should have two cubes much smaller, namely, $f^3$ and $g^3$, whose sum would be the double of a cube. 5. Case 2. Let us now suppose $p$ divisible by 3, and, consequently, that $q$ is not so. If we make $p=3r$, our formula becomes $3r(9r^2+3q^2)=9r(3r^2+q^2)$, and these factors being now numbers prime to one another, each must separately be a cube. 6. In order therefore to transform the second $q^2+3r^2$, into a cube, we shall make $q=t(t^2-9u^2)$, and $r=3u(t^2-u^2)$; and again one of the numbers $t$ and $u$ must be odd, and the other even, since otherwise the two numbers $q$ and $r$ would be even. Now, from this we obtain the first factor $9r=27u(t^2-u^2)$; and as it must be a cube, let us divide it by 27, and the formula $u(t^2-u^2)$, or $u(t+u) \times (t-u)$, must be a cube. 7. But these three factors being prime to each other, they must all be cubes of themselves. Let us therefore suppose for the last two $t+u=f^3$, and $t-u=g^3$, we shall then have $2u=f^3-g^3$; but as $u$ must be a cube, we should in this way have two cubes, in much smaller numbers, whose difference would be equal to the double of another cube. 8. Since therefore we cannot assign, in small numbers, any cubes, whose sum or difference is the double of a cube, it is evident that there are no such cubes, even among the greatest numbers. 9. It will perhaps be objected, that our conclusion might lead to error; because there does exist a satisfactory case among these small numbers; namely, that of $f=g$. But it must be considered that when $f=g$, we have, in the first case, $t+3u=t-3u$, and therefore $u=0$; consequently, also $q=0$; and, as we have supposed $x=p+q$, and $y=p-q$, the first two cubes, $x^3$ and $y^3$, must have already been equal to one another, which case was expressly excepted. Likewise, in the second case, if $f=g$, we must have $t+u=t-u$, and also $u=0$: therefore $r=0$, and $p=0$; so that the first two cubes, $x^3$ and $y^3$, would again