Elements of Algebra

410 ELEMENTS PART II. which happens with regard to those that we shall now point out. 1. It is not possible for the formula $x^4 + 4y^4$ to become a square; for since this formula is the sum of two squares, we must have $x^2 = p^2 - q^2$, and $2y^2 = 2pq$, or $y^2 = pq$; now $p$ and $q$ being numbers prime to each other, each of them must be a $\square$. If we therefore make $p = r^2$, and $q = s^2$, we shall have $x^2 = r^4 - s^4$; that is to say, the difference of two biquadrates must be a square, which is impossible. 2. Nor is it possible for the formula $x^4 - 4y^4$ to become a square; for in this case we must make $x^2 = p^2 + q^2$, and $2y^2 = 2pq$, that we may have $x^4 - 4y^4 = (p^2 - q^2)^2$; but, in order that $y^2 = pq$, both $p$ and $q$ must be squares: and if we therefore make $p = r^2$, and $q = s^2$, we have $x^2 = r^4 + s^4$; that is to say, the sum of two biquadrates must be reducible to a square, which is impossible. 3. It is impossible also for the formula $4x^4 - y^4$ to become a square, because in this case $y$ must necessarily be an even number. Now, if we make $y = 2z$, we conclude that $4x^4 - 16z^4$, and consequently, also, its fourth part, $x^4 - 4z^4$, must be reducible to a square; which we have just seen is impossible. 4. The formula $2x^4 + 2y^4$ cannot be transformed into a square; for since that square would necessarily be even, and consequently, $2x^4 + 2y^4 = 4z^2$, we should have $x^4 + y^4 = 2z^2$, or $2z^2 + 2x^2y^2 = x^4 + 2x^2y^2 + y^4 = \square$; or, in like manner, $2z^2 - 2x^2y^2 = x^4 - 2x^2y^2 + y^4 = \square$. So that, as both $2z^2 + 2x^2y^2$, and $2z^2 - 2x^2y^2$, would become squares, their product, $4z^4 - 4x^4y^4$, as well as the fourth of that product, or $z^4 - x^4y^4$, must be a square. But this last is the difference of two biquadratics; and is therefore impossible. 5. Lastly, I say also that the formula $2x^4 - 2y^4$ cannot be a square; for the two numbers $x$ and $y$ cannot both be even, since, if they were, they would have a common divisor; nor can they be the one even and the other odd, because then one part of the formula would be divisible by 4, and the other only by 2; and thus the whole formula would only be divisible by 2; therefore these numbers $x$ and $y$ must both be odd. Now, if we make $x=p+q$, and $y=p-q$, one of the numbers $p$ and $q$ will be even and the other will be odd; and, since $2x^4 - 2y^4 = 2(x^2+y^2) \times (x^2-y^2)$, and $x^2+y^2 = 2p^2+2q^2 = 2(p^2+q^2)$, and $x^2-y^2 = 4pq$, our formula will be expressed by $16pq(p^2+q^2)$, the sixteenth part of which, or $pq(p^2+q^2)$, must likewise be a square. But these factors are prime to each other, so that each of