Elements of Algebra

CHAP. XIV. OF ALGEBRA. 441 $4f^2g^2 \times h^2k^2$, by which means it becomes equal to $fg(f^2 + g^2) \times hk(h^2 + k^2)$. Lastly, this formula becomes precisely the same as that before found, if we make $f=a+b$, $g=a-b$, $h=c+d$, and $k=c-d$; since we have then $2(a^4-b^4) \times 2(c^4-d^4) = 4 \times (a^4-b^4) \times (c^4-d^4)$, which takes place, as we have seen, when $a^2=9$, $b^2=4$, $c^2=81$, and $d^2=49$; or $a=3$, $b=2$, $c=9$, and $d=7$. Thus, $f=5$, $g=1$, $h=16$, and $k=2$, whence $pq = \frac{13}{5}$, and $\frac{q}{p} = \frac{260}{64} = \frac{65}{16}$; the product of these two equations gives $q^2 = \frac{65 \times 13}{16 \times 5} = \frac{13 \times 13}{16}$; wherefore $q = \frac{13}{4}$, and it fol- lows that $p = \frac{4}{5}$, by which means we have $\frac{x}{z} = \frac{p^2+1}{p^2-1} = -\frac{41}{9}$, and $\frac{y}{z} = \frac{q^2+1}{q^2-1} = \frac{185}{153}$; therefore, since $x = -\frac{41z}{9}$, and $y = \frac{185z}{153}$, in order to obtain whole numbers, let us make $z=153$, and we shall have $x=-697$, and $y=185$. Consequently, the three square numbers sought are, $\left\{ \begin{aligned} x^2 &= 485809 \\ y^2 &= 34225 \\ z^2 &= 23409 \end{aligned} \right.$ and $\left\{ \begin{aligned} x^2 - y^2 &= 451584 = (672)^2 \\ y^2 - z^2 &= 10816 = (104)^2 \\ x^2 - z^2 &= 462400 = (680)^2 \end{aligned} \right.$. It is farther evident, that these squares are much less than those which we should have found, by squaring the three numbers $x$, $y$, and $z$ of the preceding solution. 237. Without doubt it will here be objected, that this solution has been found merely by trial, since we have made use of the Table in Article 235. But in reality we have only made use of this, to get the least possible numbers; for if we were indifferent with regard to brevity in the calcula- tion, it would be easy, by means of the rules above given, to find an infinite number of solutions; because, having found $\frac{x}{z} = \frac{p^2+1}{p^2-1}$, and $\frac{y}{z} = \frac{q^2+1}{q^2-1}$, we have reduced the question to that of transforming the product $(p^2q^2-1) \times (\frac{q^2}{p^2}-1)$ into a square. If we therefore make $\frac{q}{p} = m$, or $q=mp$, our formula will become $(m^2p^4-1) \times (m^2-1)$, which is evidently a square, when $p=1$; but we shall farther see,