Elements of Algebra

432 ELEMENTS PART II. 2. If $p = 4$, we shall have $p^2 - 1 = 15$. Now, if $x = 5$, and $y = 3$, we find $z = 16$, or $z = 0$; wherefore the three numbers sought are 3, 5, and 16. 3. If $p = 5$, we shall have $p^2 - 1 = 24$; and if we farther make $x = 3$, and $y = 8$, we find $z = 21$, or $z = 1$; whence the following numbers result; 1, 3, and 8; or 3, 8, and 21. 232. Question 11. Required three whole numbers $x, y$, and $z$, such, that if we add a given number, $a$, to each product of these numbers, multiplied two and two, we may obtain a square each time. Here we must make squares of the three following formulæ, $xy + a$, $xz + a$, and $yz + a$. Let us therefore suppose the first $xy + a = p^2$, and make $z = x + y + q$; then we shall have, for the second formula, $x^2 + xy + xq + a = x^2 + xq + p^2$; and, for the third, $xy + y^2 + yq + a = y^2 + qy + p^2$; and these both become squares by making $q = \pm 2p$: so that $z = x + y \pm 2p$; that is to say, we may find two different values for $z$. 233. Question 12. Required four whole numbers, $x, y, z$, and $v$, such, that if we add a given number, $a$, to the products of these numbers, multiplied two by two, each of the sums may be a square. Here, the six following formulæ must become squares: 1. $xy + a$, 2. $xz + a$, 3. $yz + a$, 4. $xv + a$, 5. $yv + a$, 6. $zv + a$. If we begin by supposing the first $xy + a = p^2$, and take $z = x + y + 2p$, the second and third formulæ will become squares. If we farther suppose $v = x + y - 2p$, the fourth and fifth formulæ will likewise become squares; there remains therefore only the sixth formula, which will be $x^2 + 2xy + y^2 - 4p^2 + a$, and which must also become a square. Now, as $p^2 = xy + a$, this last formula becomes $x^2 - 2xy + y^2 - 3a$; and, consequently, it is required to transform into squares the two following formulæ: $xy + a = p^2$, and $(x - y)^2 - 3a$. If the root of the last be $(x - y) - q$, we shall have $(x - y)^2 - 3a = (x - y)^2 - 2q(x - y) + q^2$; so that $-3a = -2q(x - y) + q^2$, and $x - y = \frac{q^2 + 3a}{2q}$, or $x = y + \frac{q^2 + 3a}{2q}$; consequently, $p^2 = y^2 + \frac{q^2 + 3a}{2q}y + a$. If $p = y + r$, we shall have