Elements of Algebra

CHAP. XIV. OF ALGEBRA. 437 1. $x + y = q^2 + r^2 + 2z$; 2. $x + z = q^2 + 2z$; 3. $y + z = r^2 + 2z$. Let the first $q^2 + r^2 + 2z, = t^2$, by which means $2z = t^2 - q^2 - r^2$; we must also have $t^2 - r^2 = \square$, and $t^2 - q^2 = \square$; that is to say, $t^2 - (a^2 - b^2)^2 = \square$, and $t^2 - (a^2 + b^2)^2 = \square$; we shall have to consider the two formulae $t^2 - a^4 - b^4 + 2a^2b^2$, and $t^2 - a^4 - b^4 - 2a^2b^2$. Now, as both $c^2 + d^2 + 2cd$, and $c^2 + d^2 - 2cd$, are squares, it is evident that we shall obtain what we want by comparing $t^2 - a^4 - b^4$, with $c^2 + d^2$, and $2a^2b^2$ with $2cd$. With this view, let us suppose $cd = a^2b^2 = f^2g^2h^2k^2$, and take $c = f^2g^2$, and $d = h^2k^2$; $a^2 = f^2h^2$, and $b^2 = g^2k^2$, or $a = fh$, and $b = gk$; the first equation $t^2 - a^4 - b^4 = c^2 + d^2$, will assume the form $t^2 - f^4h^4 - g^4k^4 = f^4g^4 + h^4k^4$; whence $t^2 = f^4g^4 + f^4h^4 + g^4k^4 + h^4k^4$, or $t^2 = (f^4 + k^4) \times (g^4 + h^4)$; consequently, this product must be a square; but as the resolution of it would be difficult, let us consider the subject under a different point of view. If from the first three equations $x - y = p^2$, $x - z = q^2$, $y - z = r^2$, we determine the letters $y$ and $z$, we shall find $y = x - p^2$, and $z = x - q^2$; whence it follows that $q^2 = p^2 + r^2$. Our first formulae now become $x+y = 2x - p^2$, $x+z = 2x - q^2$, and $y+z = 2x - p^2 - q^2$. Let us make this last $2x - p^2 - q^2 = t^2$, so that $2x = t^2 + p^2 + q^2$, and there will only remain the formulae $t^2 + q^2$, and $t^2 + p^2$, to transform into squares. But since we must have $q^2 = p^2 + r^2$, let $q = a^2 + b^2$, and $p = a^2 - b^2$; and we shall then have $r = 2ab$; consequently, our formulae will be: 1. $t^2 + (a^2 + b^2)^2 = t^2 + a^4 + b^4 + 2a^2b^2 = \square$; 2. $t^2 + (a^2 - b^2)^2 = t^2 + a^4 + b^4 - 2a^2b^2 = \square$. In order to accomplish our purpose, we have only to compare again $t^2 + a^4 + b^4$ with $c^2 + d^2$, and $2a^2b^2$ with $2cd$. Therefore, as before, let $c = f^2g^2$, $d = h^2k^2$, $a = fh$, and $b = gk$; we shall then have $cd = a^2b^2$, and we must again have $t^2 + f^4h^4 + g^4k^4 = c^2 + d^2 = f^4g^4 + h^4k^4$; whence $t^2 = f^4g^4 - f^4h^4 + h^4k^4 - g^4k^4 = (f^4 - k^4) \times (g^4 - h^4)$. So that the whole is reduced to finding the differences of two pair of biquadrates, namely, $f^4 - k^4$, and $g^4 - h^4$, which, multiplied together, may produce a square. For this purpose, let us consider the formula $m^4 - n^4$; let us see what numbers it furnishes, if we substitute given numbers for $m$ and $n$, and attend to the squares that will be