Elements of Algebra

CHAP. XII. OF ALGEBRA. 401 longer exist; because the two factors, which would then be $x+y$, and $x-y$, might have common divisors, even when $x$ and $y$ had none; as would be the case, for example, if both these letters expressed odd numbers. Thus, when $x^2-y^2$ must become a cube, it is not necessary that both $x+y$, and $x-y$, should of themselves be cubes; but we may suppose $x+y=2p^3$, and $x-y=4q^3$; and the formula $x^2-y^2$ will undoubtedly become a cube, since we shall find it to be $8p^3q^3$, the cube root of which is $2pq$. We shall farther have $x=p^3+2q^3$, and $y=p^3-2q^3$. On the contrary, when the formula $ax^2+cy^2$ is not resolvible into two rational factors, we cannot find any other solutions beside those which have been already given. 192. We shall illustrate the preceding investigations by some curious examples. Question 1. Required a square, $x^2$, in integer numbers, and such, that, by adding 4 to it, the sum may be a cube. The condition is answered when $x^2=121$; but we wish to know if there are other similar cases. As 4 is a square, we shall first seek the cases in which $x^2+y^2$ becomes a cube. Now, we have found one case, namely, if $x=p^3-3pq^2$, and $y=3p^2q-q^3$: therefore, since $y^2=4$, we have $y=\pm 2$, and, consequently, either $3p^2q-q^3=+2$, or $3p^2q-q^3=-2$. In the first case, we have $q(3p^2-q^2)=2$, so that $q$ is a divisor of 2. This being laid down, let us first suppose $q=1$, and we shall have $3p^2-1=2$; therefore $p=1$; whence $x=2$, and $x^2=4$. If, in the second place, we suppose $q=2$, we have $6p^2-8=\pm 2$; admitting the sign $+$, we find $6p^2=10$, and $p^2=\frac{5}{3}$; whence we should get an irrational value of $p$, which could not apply here; but if we consider the sign $-$, we have $6p^2=6$, and $p=1$; therefore $x=11$: and these are the only possible cases; so that 4, and 121, are the only two squares, which, added to 4, give cubes. 193. Question 2. Required, in integer numbers, other squares, beside 25, which, added to 2, give cubes. Since $x^2+2$ must become a cube, and since 2 is the double of a square, let us first determine the cases in which $x^2+2y^2$ becomes a cube; for which purpose we have, by Article 188, in which $a=1$, and $c=2$, $x=p^3-6pq^2$, and $y=3p^2q-2q^3$: therefore, since $y=\pm 1$, we must have $3p^2q-q^3$, or $q(3p^2-2q^2)=\pm 1$; and, consequently, $q$ must be a divisor of 1. Therefore let $q=1$, and we shall have $3p^2-2=\pm 1$. DD