Elements of Algebra

452 ELEMENTS PART II. sequently, the eighth part of our formula, or $\frac{1}{4}p(p^2 + 3q^2)$, would necessarily be a whole number, and also a cube. Now, we know that one of the numbers $p$ and $q$ is even, and the other odd; so that $p^2+3q^2$ must be an odd number, which not being divisible by 4, $p$ must be so, or $\frac{p}{4}$ must be a whole number. 5. But in order that the product $\frac{1}{4}p(p^2+3q^2)$ may be a cube, each of these factors, unless they have a common divisor, must separately be a cube; for if a product of two factors, that are prime to each other, be a cube, each of itself must necessarily be a cube; and if these factors have a common divisor, the case is different, and requires a particular consideration. So that the question here is, to know if the factors $p$, and $p^2+3q^2$, might not have a common divisor. To determine this, it must be considered, that if these factors have a common divisor, the numbers $p^2$, and $p^2+3q^2$, will have the same divisor; that the difference also of these numbers, which is $3q^2$, will have the same common divisor with $p^2$; and that, since $p$ and $q$ are prime to each other, these numbers $p^2$, and $3q^2$, can have no other common divisor than 3, which is the case when $p$ is divisible by 3. 6. We have consequently two cases to examine: the one is, that in which the factors $p$, and $p^2+3q^2$, have no common divisor, which happens always, when $p$ is not divisible by 3; the other case is, when these factors have a common divisor, and that is when $p$ may be divided by 3; because then the two numbers are divisible by 3. We must carefully distinguish these two cases from each other, because each requires a particular demonstration. 7. Case 1. Suppose that $p$ is not divisible by 3, and, consequently, that our two factors $\frac{p}{4}$, and $p^2+3q^2$, are prime to each other; so that each must separately be a cube. Now, in order that $p^2+3q^2$ may become a cube, we have only, as we have seen before, to suppose $p+q\sqrt{-3}=(t+u\sqrt{-3})^3$, and $p-q\sqrt{-3}=(t-u\sqrt{-3})^3$, which gives $p^2+3q^2=(t^2+3u^2)^3$, which is a cube, and gives us $p=t^3-9tu^2=t(t^2-9u^2)$, also $q=3t^2u-3u^3=3u(t^2-u^2)$. Since therefore $q$ is an odd number, $u$ must also be odd; and, consequently, $t$ must be even, because otherwise $t^2-u^2$ would be even. 8. Having transformed $p^2+3q^2$ into a cube, and having