Elements of Algebra

CHAP. XIV. OF ALGEBRA. 435 let us take $b/c = 3/2$, and $d/e = 4/3$, we shall then have $f/g = 7/5$; whence $x = 25 - 9a$, $y = 49 - 16a$, and $z = 144 - 49a$; by which means we have 1. $xy + a = 1225 - 840a + 144a^2 = (35 - 12a)^2$; 2. $xz + a = 3600 - 2520a + 441a^2 = (60 - 21a)^2$; 3. $yz + a = 7056 - 4704a + 784a^2 = (84 - 28a)^2$. 234. In order now to determine, according to our question, four letters, $x, y, z$, and $v$, we must add a fourth fraction to the three preceding: therefore let the first three be $b/c$, $d/e$, $f/g = (b+d)/(c+e)$, and let us suppose the fourth fraction $h/k = (b+d)/(e+g) = (2d+b)/(2e+c)$, so that it may have the given relation with the third and second; if after this we make $x = b^2 - ac^2$, $y = d^2 - ae^2$, $z = f^2 - ag^2$, and $v = h^2 - ak^2$, we shall have already fulfilled the following conditions: $xy + a = \square$, $xz + a = \square$, $yz + a = \square$, $yv + a = \square$, $zy + a = \square$. It therefore only remains to make $xv + a$ become a square, which does not result from the preceding conditions, because the first fraction has not the necessary relation with the fourth. This obliges us to preserve the indeterminate number $m$ in the three first fractions; by means of which, and by determining $m$, we shall be able also to transform the formula $xv + a$ into a square. 6. If we therefore take the first case from our small Table, and make $b/c = 3/2$, and $d/e = (3m+1)/(2m+1)$; we shall have $f/g = (3m+4)/(2m+3)$, and $h/k = (6m+5)/(4m+4)$, whence $x = 9 - 4a$, and $v = (6m+5)^2 - a(4m+4)^2$; so that $xv + a = 9(6m+5)^2 - 4a(6m+5)^2 - 9a(4m+4)^2 + 4a^2(4m+4)^2 + a$; or $xv + a = 9(6m+5)^2 + 4a^2(4m+4)^2 - a(288m^2 + 528m + 244)$, which we can easily transform into a square, since $m^2$ will be found to be multiplied by a square; but on this we shall not dwell. 7. The fractions, which have been found to be neces- FF2