Elements of Algebra

444 ELEMENTS PART II. after being multiplied by $4p^2q^2$; that is, we must have $q^2(p^2-1)^2 + p^2(q^2-1)^2 = \square$. Now, this can scarcely be obtained, unless we previously know a case in which this formula becomes a square: and as it is also difficult to find such a case, we must have recourse to other artifices, some of which we shall now explain. 1. As the formula in question may be expressed thus, $q^2(p+1)^2 \times (p-1)^2 + p^2(q+1)^2 \times (q-1)^2 = \square$, let us make it divisible by the square $(p+1)^2$; which may be done by making $q-1=p+1$, or $q=p+2$; for then $q+1=p+3$, and the formula becomes $(p+2)^2 \times (p+1)^2 \times (p-1)^2 + p^2(p+3)^2 \times (p+1)^2 = \square$; so that dividing by $(p+1)^2$, we have $(p+2)^2 \times (p-1)^2 + p^2(p+3)^2$, which must be a square, and to which we may give the form $2p^4 + 8p^3 + 6p^2 - 4p + 4$. Now, the last term here being a square, let us suppose the root of the formula to be $2+fp+gp^2$, or $gp^2+fp+2$, the square of which is $g^2p^4 + 2fgp^3 + 4gp^2 + f^2p^2 + 4fp + 4$, and we shall destroy the last three terms, by making $4f=-4$, or $f=-1$, and $4g+1=6$, or $g=\frac{5}{4}$. Also the first terms being divided by $p^3$, will give $2p+8=g^2p+2fg=\frac{25}{16}p-\frac{5}{2}$; or $p=-24$, and $q=-22$; whence $\frac{x}{z} = \frac{p^2-1}{2p} = -\frac{575}{48}$, or $x = -\frac{575}{48}z$, and $\frac{y}{z} = \frac{q^2-1}{2q} = -\frac{483}{44}$, or $y = -\frac{483}{44}z$. Let us now make $z = 16 \times 3 \times 11$; we shall then have $x = 575 \times 11$, and $y = 483 \times 12$; consequently, the roots of the three squares sought will be: $x = 6325 = 11 \times 23 \times 25$; $y = 5796 = 12 \times 21 \times 23$; and $z = 528 = 3 \times 11 \times 16$; for from these result, $x^2 + y^2 = 23^2(275^2 + 252^2) = 23^2 \times 373^2$; $x^2 + z^2 = 11^2(575^2 + 48^2) = 11^2 \times 577^2$; and $y^2 + z^2 = 12^2(483^2 + 44^2) = 12^2 \times 485^2$. 2. We may also make our formula divisible by a square, in an infinite number of ways; for example, if we suppose $(q+1)^2 = 4(p+1)^2$, or $q+1=2(p+1)$, that is to say, $q=2p+1$, and $q-1=2p$, the formula will become $(2p+1)^2 \times (p+1)^2 \times (p-1)^2 + p^2 \times 4(p+1)^2 \times 4p^2 = \square$; which may be divided by $(p+1)^2$, by which means we have $(2p+1)^2 \times (p-1)^2 + 16p^4 = \square$, or $20p^4 - 4p^3 - 3p^2 + 2p + 1 = \square$; but from this we derive nothing.