Elements of Algebra

CHAP. XIV. OF ALGEBRA. 415 Let $u=3$, and $t=2$; we shall then have $x=\frac{120}{169}$; and the formulæ $1+x=\frac{289}{169}$, and $1-x=\frac{49}{169}$, will both be squares. 216. Question 4. To find such numbers $x$, that whether they be added to 10, or subtracted from 10, the sum and the difference may be squares. It is required, therefore, to transform into squares the formulæ $10+x$, and $10-x$, which might be done by the method that has just been employed; but let us explain another mode of proceeding. It will be immediately perceived, that the product of these two formulæ, or $100-x^2$, must likewise become a square. Now, its first term being already a square, we may suppose its root to be $10-px$, by which means we shall have $100-x^2=100-20px+p^2x^2$; therefore $p^2x+x=20p$, and $x=\frac{20p}{p^2+1}$. Now, from this it is only the product of the two formulæ which becomes a square, and not each of them separately: but provided one becomes a square, the other will necessarily be also a square. Now $10+x=\frac{10p^2+20p+10}{p^2+1}=\frac{10(p^2+2p+1)}{p^2+1}$, and since $p^2+2p+1$ is already a square, the whole is reduced to making the fraction $\frac{10}{p^2+1}$, or $\frac{10p^2+10}{(p^2+1)^2}$, a square also. For this purpose we have only to make $10p^2+10$ a square, and here it is necessary to find a case in which that takes place. It will be perceived that $p=3$ is such a case; for which reason we shall make $p=3+q$, and shall have $100+60q+10q^2$. Let the root of this be $10+qt$, and we shall have the final equation, $100+60q+10q^2=100+20qt+q^2t^2$, which gives $q=\frac{60-20t}{t^2-10}$, by which means we shall determine $p=3+q$, and $x=\frac{20p}{p^2+1}$. Let $t=3$, we shall then find $q=0$, and $p=3$; therefore $x=6$, and our formulæ $10+x=16$, and $10-x=4$. But if $t=1$, we have $q=-\frac{40}{9}$, and $p=-\frac{13}{9}$, so that $x=-\frac{234}{25}$; now it is of no consequence if we also make $x=+\frac{234}{25}$; therefore $10+x=\frac{484}{25}$, and $10-x=\frac{16}{25}$, which quantities are both squares. 217. *Remark*. If we wished to generalise this question,