Elements of Algebra

420 ELEMENTS PART II. A greater number of values will be found for $q$, by making use of those which we have already determined. Thus, having found $q = -\frac{1}{2}$; let now $q = -\frac{1}{2} + r$, and we shall have $p = \frac{1}{2} + r$; also, $p^2 = \frac{1}{4} + r + r^2$, and $p^4 = \frac{1}{16} + \frac{1}{2}r + \frac{3}{2}r^2 + 2r^3 + r^4$; whence the expression $p^4 - 2p^2 + 2 = \frac{25}{16} - \frac{3}{2}r - \frac{1}{2}r^2 + 2r^3 + r^4$, to which our formula, $x^2 + 1$, is reduced, must be a square, and it must also be so when multiplied by 16; in which case, we have $25 - 24r - 8r^2 + 32r^3 + 16r^4$ to be a square. For which reason, let us now represent 1. The root by $5 + fr \pm 4r^2$; so that $25 - 24r - 8r^2 + 32r^3 + 16r^4 =$ $25 + 10fr \pm 40r^2 + f^2r^2 \pm 8fr^3 + 16r^4$. The first and the last terms destroy each other; and we may destroy the second also, if we make $10f = -24$, and, consequently, $f = -\frac{12}{5}$; then dividing the remaining terms by $r^2$, we have $-8 + 32r = \pm 40 + f^2 \pm 8fr$; and, admitting the upper sign, we find $r = \frac{48+f^2}{32-8f}$. Now, because $f = -\frac{12}{5}$, we have $r = \frac{21}{20}$; therefore $p = \frac{31}{20}$, and $x = \frac{561}{400}$; so that $x+1 = (\frac{31}{20})^2$, and $x^2+1 = (\frac{689}{400})^2$. 2. If we adopt the lower sign, we have $-8 + 32r = -40 + f^2 - 8fr$, whence $r = \frac{f^2-32}{32+8f}$; and since $f = -\frac{12}{5}$, we have $r = -\frac{41}{40}$; therefore $p = \frac{31}{40}$, which leads to the preceding equation. 3. Let $4r^2 + 4r \pm 5$ be the root; so that $16r^4 + 32r^3 - 8r^2 - 24r + 25 =$ $16r^4 + 32r^3 \pm 40r^2 + 16r^2 \pm 40r + 25$: and as on both sides the first two terms and the last destroy each other, we shall have $-8r^2 - 24r = \pm 40r^2 + 16r^2 \pm 40r$, or $-24r - 24 = \pm 40r \pm 40$. Here, if we admit the upper sign, we shall have $-24r - 24 = 40r + 40$, or $0 = 64r + 64$, or $0 = r+1$, that is, $r = -1$, and $p = -\frac{1}{2}$; but this is a case already known to us, and we should not have found a different one by making use of the other sign. 4. Let now the root be $5 + fr + gr^2$, and let us determine $f$ and $g$ so, that the first three terms may vanish: then, since