Elements of Algebra

CHAP. XIV. OF ALGEBRA. 431 231. Question 10. To determine three numbers, $x$, $y$, and $z$, such, that multiplying them together two and two, and adding 1 to the product, we may obtain a square each time; that is, to transform into squares the three following formulæ: $xy + 1$, $xz + 1$, and $yz + 1$. Let us suppose one of the last two, as $xz + 1, = p^2$, and the other $yz + 1 = q^2$, and we shall have $x = \frac{p^2-1}{z}$, and $y = \frac{q^2-1}{z}$. The first formula is now transformed to $\frac{(p^2-1) \times (q^2-1)}{z^2} + 1$; which must consequently be a square, and will be no less so, if multiplied by $z^2$; so that $(p^2-1) \times (q^2-1) + z^2$, must be a square, which it is easy to form. For, let its root be $z+r$, and we shall have $(p^2-1) \times (q^2-1) = 2rz + r^2$, and $z = \frac{(p^2-1) \times (q^2-1) - r^2}{2r}$, in which any numbers may be substituted for $p$, $q$, and $r$. For example, if $r = (pq+1)$, we shall have $r^2 = p^2q^2 + 2qp + 1$, and $z = \frac{p^2+2pq+q^2}{2pq+2}$; wherefore $x = \frac{(p^2-1) \times (2pq+2)}{p^2+2pq+q^2} = \frac{2(pq+1) \times (p^2-1)}{(p+q)^2}$, and $y = \frac{2(pq+1) \times (q^2-1)}{(p+q)^2}$. But if whole numbers be required, we must make the first formula $xy + 1 = p^2$, and suppose $z = x + y + q$; then the second formula becomes $x^2 + xy + xq + 1 = x^2 + qx + p^2$, and the third will be $xy + y^2 + qy + 1 = y^2 + qy + p^2$. Now, these evidently become squares, if we make $q = \pm 2p$; since in that case the second is $x^2 \pm 2px + p^2$, the root of which is $x \pm p$, and the third is $y^2 \pm 2py + p^2$, the root of which is $y \pm p$. We have consequently this very elegant solution: $xy+1=p^2$, or $xy = p^2 - 1$, which applies easily to any value of $p$; and from this the third number also is found, in two ways, since we have either $z = x + y + 2p$, or $z = x + y - 2p$. Let us illustrate these results by some examples. 1. Let $p = 3$, and we shall have $p^2 - 1 = 8$; if we make $x=2$, and $y=4$, we shall have either $z=12$, or $z=0$; so that the three numbers sought are 2, 4, and 12.