Elements of Algebra

CHAP. XIII. OF ALGEBRA. 411 them must be a square. Let us, therefore, make the first two $p = r^2$, and $q = s^2$, and the third will become $r^4 + s^4$, which cannot be a square, therefore the given formula cannot become a square. 210. We may likewise demonstrate, that the formula $x^4 + 2y^4$ can never become a square: the rationale of this demonstration being as follows: 1. The number $x$ cannot be even, because in that case $y$ must be odd; and the formula would only be divisible by $2$, and not by $4$; so that $x$ must be odd. 2. If, therefore, we suppose the square root of our formula to be $x^2 + \frac{2py^2}{q}$, in order that it may become odd, we shall have $x^4 + 2y^4 = x^4 + \frac{4px^2y^2}{q} + \frac{4p^2y^4}{q^2}$, in which the terms $x^4$ are destroyed; so that if we divide the other terms by $y^2$, and multiply by $q^2$, we find $4pqx^2 + 4p^2y^2 = 2q^2y^2$, or $4pqx^2 = 2q^2y^2 - 4p^2y^2$, whence we obtain $\frac{x^2}{y^2} = \frac{q^2 - 2p^2}{2pq}$; that is, $x^2 = q^2 - 2p^2$, and $y^2 = 2pq$*, which are the same formulæ that have been already given. 3. So that $q^2 - 2p^2$ must be a square, which cannot happen, unless we make $q = r^2 + 2s^2$, and $p = 2rs$, in order to have $x^2=(r^2 - 2s^2)^2$; now, this will give us $4rs(r^2 + 2s^2) = y^2$; and its fourth part, $rs(r^2 + 2s^2)$ must also be a square: consequently $r$ and $s$ must respectively be each a square. If, therefore, we suppose $r = t^2$, and $s = u^2$, we shall find the third factor $r^2 + 2s^2 = t^4 + 2u^4$, which ought to be a square. 4. Consequently, if $x^4 + 2y^4$ were a square, $t^4 + 2u^4$ must also be a square; and as the numbers $t$ and $u$ would be much less than $x$ and $y$, we should always come, in the same manner, to numbers successively less: but as it is easy from trials to be convinced, that the given formula is not a square in any small number; it cannot therefore be the square of a very great number. 211. On the contrary, with regard to the formula $x^4 - 2y^4$, it is impossible to prove that it cannot become a square; and, by a process of reasoning similar to the foregoing, we even find that there are an infinite number of cases in which this formula really becomes a square. In fact, if $x^4 - 2y^4$ must become a square, we shall see * Because $x$ and $y$ are prime to each other.