Elements of Algebra

CHAP. XIV. OF ALGEBRA. 421 $\frac{25 - 24r - 8r^2 + 32r^3 + 16r^4}{25 + 10fr + 10f^2r^2 + 10gr^2 + 2fgr^3 + g^2r^4}$, we shall first have $10f = -24$, so that $f = -\frac{12}{5}$; then $10g + f^2 = -8$, or $g = \frac{-8-f^2}{10} = \frac{-344}{250} = \frac{-172}{125}$. When, therefore, we have substituted and divided the remaining terms by $r^3$, we shall have $32 + 16r = 2fg + g^2r$, and $r = \frac{2fg - 32}{16 - g^2}$. Now, the numerator $2fg - 32$ becomes here $\frac{24 \times 172 - 32 \times 625}{5 \times 125} = \frac{-32 \times 496}{625} = \frac{-16 \times 32 \times 31}{625}$, and the denominator $16 - g^2 = (4-g) \times (4+g) = \frac{328}{125} \times \frac{672}{125} = \frac{8 \times 32 \times 41 \times 21}{25 \times 625}$; so that $r = -\frac{1550}{861}$; and hence we conclude that $p = -\frac{2239}{1722}$; by means of which we obtain a new value of $x$, because $x = p^2 - 1$. 223. Question 8. To find a number, $x$, which, added to each of the given numbers, $a, b, c$, produces a square. Since here the three formulæ $x+a$, $x+b$, and $x+c$, must be squares, let us make the first $x+a = z^2$, and we shall have $x = z^2 - a$, and the two other formulæ will, by substitution, be changed into $z^2 + b - a$, and $z^2 + c - a$. It is now required for each of these to be a square; but this does not admit of a general solution: the problem is frequently impossible, and its possibility entirely depends on the nature of the numbers $b-a$, and $c-a$. For example, if $b-a=1$, and $c-a=-1$, that is to say, if $b=a+1$, and $c=a-1$, it would be required to make $z^2+1$, and $z^2-1$ squares, and, consequently, that $z$ should be a fraction; so that we should make $z = \frac{p}{q}$. It would be farther necessary that the two formulæ, $p^2+q^2$, and $p^2-q^2$, should be squares, and, consequently, that their product also, $p^4-q^4$, should be a square. Now, we have already shewn (Art. 202) that this is impossible. Were we to make $b-a=2$, and $c-a=-2$; that is, $b=a+2$, and $c=a-2$; and also, if $z = \frac{p}{q}$, we should have the two formulæ, $p^2+2q^2$, and $p^2-2q^2$, to transform into squares; consequently, it would also be necessary for