Elements of Algebra

418 ELEMENTS PART II. infinite number of other ways: for, since $p = 5$, and $q = 1$, if we write $t$ and $u$, instead of $p$ and $q$, and $p$ and $q$, instead of $x$ and $y$, in the formulæ of the foregoing example, we shall find $p = \frac{2tu+5(u^2-t^2)}{t^2+u^2}$, and $q = \frac{10tu+t^2-u^2}{t^2+u^2}$. Here we may now substitute any numbers for $t$ and $u$, and by those means determine $p$ and $q$, and, consequently, also the value of $x = \frac{p^2-q^2}{2}$. For example, let $t = 2$, and $u = 1$; we shall then have $p = \frac{13}{5}$, and $q = \frac{23}{5}$; wherefore $p^2 - q^2 = -\frac{408}{25}$, and $x = -\frac{204}{25}$. 221. But, in order to resolve this question generally, let $a = c^2 + d^2$, and put $z$ for the unknown quantity; that is to say, the formulæ, $a+z$, and $a-z$, must become squares. Let us therefore make $a+z = x^2$, and $a-z = y^2$, and we shall thus have first $2a = 2(c^2+d^2) = x^2+y^2$, then $2z = x^2-y^2$. Therefore the squares $x^2$ and $y^2$ must be such, that $x^2+y^2 = 2(c^2+d^2)$; where $2(c^2+d^2)$ is really the sum of two squares, namely, $(c+d)^2 + (c-d)^2$; and, in order to abbreviate, let us suppose $c+d=f$, and $c-d=g$; then we must have $x^2+y^2 = f^2+g^2$; and this will happen, according to what has been already said, when $x = \frac{2gpq+f(q^2-p^2)}{p^2+q^2}$, and $y = \frac{2fpq+g(p^2-q^2)}{p^2+q^2}$; from which we obtain a very easy solution, by making $p=1$, and $q=1$; for we find $x = \frac{2g}{2} = g = c-d$, and $y=f=c+d$; consequently, $z=2cd$; and it is evident that $a+z = c^2+2cd+d^2 = (c+d)^2$, and $a-z = c^2-2cd+d^2 = (c-d)^2$. Let us attempt another solution, by making $p=2$, and $q=1$; we shall then have $x = \frac{c-7d}{5}$, and $y = \frac{7c+d}{5}$, where $c$ and $d$, as well as $x$ and $y$, may be taken minus, because we have only to consider their squares. Now, since $x$ must be greater than $y$, let us make $d$ negative, and we shall have $x = \frac{c+7d}{5}$, and $y = \frac{7c-d}{5}$: hence \ldots