Elements of Algebra

CHAP. XIV. OF ALGEBRA. 445 3. Let us then rather make $(q-1)^2 = 4(p+1)^2$, or $q-1 = 2(p+1)$; we shall then have $q = 2p+3$, and $q+1 = 2p+4$, or $q+1 = 2(p+2)$, and after having divided our formula by $(p+1)^2$, we shall obtain the following; $(2p+3)^2 \times (p-1)^2 + 16p^2(p+2)^2$, or $9 - 6p + 53p^2 + 68p^3 + 20p^4$. Let its root be $3-p+gp^2$, the square of which is $9 - 6p + 6gp^2 + p^2 - 2gp^3 + g^2p^4$; the first two terms vanish, and we may destroy the third by making $6g+1 = 53$, or $g = \frac{26}{3}$; so that the other terms are divisible by $p$, and give $20p + 68 = g^2p - 2g$, or $\frac{496}{9}p = \frac{256}{3}$; therefore $p = \frac{48}{31}$, and $q = \frac{189}{31}$, by which means we obtain a new solution. 4. If we make $q-1 = \frac{4}{3}(p-1)$, we have $q = \frac{4}{3}p - \frac{1}{3}$, and $q+1 = \frac{4}{3}p + \frac{2}{3} = \frac{2}{3}(2p+1)$, and the formula, after being divided by $(p-1)^2$, becomes $(\frac{4p-1}{9})^2 \times (p+1)^2 + \frac{64}{81}p^2(2p+1)^2$; multiplying by 81, we have $9(4p-1) \times (p+1)^2 + 64p^2(2p+1)^2 = 400p^4 + 472p^3 + 73p^2 - 54p + 9$, in which the first and last terms are both squares. If, therefore, we suppose the root to be $20p^2 - 9p + 3$, the square of which is $400p^4 - 360p^3 + 120p^2 + 81p^2 - 54p + 9$, we shall have $472p + 73 = -360p + 201$; wherefore $p = \frac{2}{13}$, and $q = -\frac{5}{39}$. We might likewise have taken for the root $20p^2+9p-3$, the square of which is $400p^4+360p^3-120p^2+81p^2-54p+9$; but comparing this square with our formula, we should have found $472p+73 = 360p-39$, and consequently $p=-1$, a value which can be of no use to us. 5. We may also make our formula divisible by the two squares, $(p+1)^2$, and $(p-1)^2$, at the same time. For this purpose, let us make $q = \frac{pt+1}{p+t}$; so that $q+1 = \frac{pt+p+t+1}{p+t} = \frac{(p+1)\times(t+1)}{p+t}$, and $q-1 = \frac{pt-p-t+1}{p+t} = \frac{(p-1)\times(t-1)}{p+t}$. This formula will be divisible by $(p+1)^2 \times (p-1)^2$, and will be reduced to $\frac{(pt+1)^2}{(p+t)^2} + \frac{(t+1)^2\times(t-1)^2}{(p+t)^4} \times p^2$. If we multiply by $(p+t)^4$, the formula, as before, must be transformable into a square, and we shall have $(pt+1)^2 \times (p+t)^2 + p^2(t+1)^2 \times (t-1)^2$, or