Elements of Algebra

CHAP. XIV. OF ALGEBRA. 443 tended to; which is that in which $a$ is a square, and takes place, for example, when $m = \frac{5}{3}$; since then $a = \frac{25}{16}$. If here again, in order to abridge, we make $a = b^2$, so that our formula may be $1 + 4b^2s + 6b^2s^2 + 4b^2s^3 + b^2s^4$, we may compare it with the square of $1 + 2b^2s + b^2s^2$, that is to say, with $1 + 4b^2s + 2b^2s^2 + 4b^3s^3 + b^4s^4$; and expunging on both sides the first two terms and the last, and dividing the rest by $s^2$, we shall have $6b^2 + 4b^3s = 2b + 4b^4s$, whence $s = \frac{6b^2-2b-4b^4}{4b^3-4b^2} = \frac{3b-1-2b^3}{2b^2-2b}$; but this fraction being still divisible by $b-1$, we shall, at last, have $s = \frac{1-2b-2b^2}{2b}$, and $p = \frac{1-2b^2}{2b}$. We might also have taken $1+2bs+bs^2$ for the root of our formula; the square of this trinomial being $1+4bs+2b^2s^2+4b^2s^3+b^2s^4$, we should have destroyed the first, and the last two terms; and dividing the rest by $s$, we should have been brought to the equation $4b^2+6b^2s = 4b+2bs+4b^2s$. But as $b^2 = \frac{25}{16}$, and $b = \frac{5}{4}$, this equation would have given us $s = -2$, and $p = -1$; consequently, $p^2-1=0$, from which we could not have drawn any conclusion, since we should have had $z=0$. To return then to the former solution, which gave $p = \frac{1-2b^2}{2b}$; as $b = \frac{5}{4}$, it shews us that if $m = \frac{5}{3}$, we have $p = \frac{17}{20}$, and $q = mp = \frac{17}{12}$; consequently, $\frac{x}{z} = \frac{689}{111}$, and $\frac{y}{x} = \frac{433}{143}$. 238. Question 15. Required three square numbers such, that the sum of every two of them may be a square. Since it is required to transform the three formulæ $x^2+y^2$, $x^2+z^2$, and $y^2+z^2$ into squares, let us divide them by $z^2$, in order to have the three following, $\frac{x^2}{z^2} + \frac{y^2}{z^2} = \square$, $\frac{x^2}{z^2} + 1 = \square$, $\frac{y^2}{z^2} + 1 = \square$. The last two are answered, by making $\frac{x}{z} = \frac{p^2-1}{2p}$, and $\frac{y}{z} = \frac{q^2-1}{2q}$, which also changes the first formula into this, $\frac{(p^2-1)^2}{4p^2} + \frac{(q^2-1)^2}{4q^2}$, which ought also to continue a square