Elements of Algebra

414 ELEMENTS PART II. second will become $x+7=p^2+3$; and, as this last formula must also be a square, let its root be represented by $p+q$, and we shall have $p^2+3=p^2+2pq+q^2$; whence we obtain $p=\frac{3-q^2}{2q}$, and, consequently, $x=\frac{9-22q^2+q^4}{4q^2}$; and if we also take a fraction $\frac{r}{s}$ for $q$, we find $x=\frac{9s^4-22r^2s^2+r^4}{4r^2s^2}$, in which we may substitute for $r$ and $s$ any integer numbers whatever. If we make $r=1$, and $s=1$, we find $x=-3$; therefore $x+4=1$, and $x+7=4$. If $x$ were required to be a positive number, we might make $s=2$, and $r=1$; we should then have $x=\frac{57}{16}$; whence $x+4=\frac{121}{16}$, and $x+7=\frac{169}{16}$. If we make $s=3$, and $r=1$, we have $x=\frac{133}{36}$; whence $x+4=\frac{169}{36}$, and $x+7=\frac{196}{36}$. In order that the last term of the formula, which expresses $x$, may exceed the middle term, let us make $r=5$, and $s=1$, and we shall have $x=\frac{21}{25}$; consequently $x+4=\frac{121}{25}$, and $x+7=\frac{196}{25}$. 215. Question 3. Required such a fractional value of $x$, that if added to 1, or subtracted from 1, it may give in both cases a square. Since the two formulæ $1+x$, and $1-x$, must become squares, let us suppose the first $1+x=p^2$, and we shall have $x=p^2-1$; also, the second formula will then be $1-x=2-p^2$. As this last formula must become a square, and neither the first nor the last term is a square, we must endeavour to find a case, in which the formula does become a $\square$, and we soon perceive one, namely, when $p=1$. If we therefore make $p=1-q$, so that $x=q^2-2q$, we have $2-p^2=1+2q-q^2$; and supposing its root to be $1-qr$, we shall have $1+2q-q^2=1-2qr+q^2r^2$; so that $2-q=-2r+qr^2$, and $q=\frac{2r+2}{r^2+1}$; whence results $x=\frac{4r-4r^3}{(r^2+1)^2}$; and since $r$ is a fraction, if we make $r=\frac{t}{u}$, we shall have $x=\frac{4tu^3-4t^3u}{(t^2+u^2)^2} = \frac{4tu(u^2-t^2)}{(t^2+u^2)^2}$, where it is evident that $u$ must be greater than $t$. Let therefore $u=2$, and $t=1$, and we shall find $x=\frac{24}{25}$.