Elements of Algebra

446 ELEMENTS PART II. $t^2 p^4 + 2t(t^2+1)p^3 + 2t^2 p^2 + (t^2+1)^2 p^2 + (t^2-1)^2 p + 2t(t^2+1)p + t^2$ in which the first and the last terms are squares. Let us therefore take for the root $tp^2 + (t^2+1)p - t$, the square of which is $t^2 p^4 + 2t(t^2+1)p^3 - 2t^2 p^2 + (t^2+1)^2 p^2 - 2t(t^2+1)p + t^2$, and we shall have, by comparing, $2t^2 p + (t^2+1)^2 p + 2t(t^2+1) + (t^2-1)^2 p = -2t^2 p + (t^2+1)^2 p - 2t(t^2+1)$; or, by subtraction, $4t^2 p + 4t(t^2+1) + (t^2-1)^2 p = 0$, or $(t^2+1)^2 p + 4t(t^2+1) = 0$, that is to say, $t^2+1 = \frac{-4t}{p}$; whence $p = \frac{-4t}{t^2+1}$; consequently, $pt+1 = \frac{-3t^2+1}{t^2+1}$, and $p+t = \frac{t^3-3t}{t^2+1}$; lastly, $q = \frac{-3t^2+1}{t^3-3t}$, where the value of the letter $t$ is arbitrary. For example, let $t=2$; we shall then have $p = -\frac{8}{5}$ and $q = -\frac{11}{2}$; so that $\frac{x}{z} = \frac{p^2-1}{2p} = +\frac{39}{80}$, and $\frac{y}{z} = \frac{q^2-1}{2q} = \frac{117}{44}$, or $x = \frac{3 \times 13}{4 \times 4 \times 5}z$, and $y = \frac{9 \times 13}{4 \times 11}z$. Farther, if $x = 3 \times 11 \times 13$, we have $y = 4 \times 5 \times 9 \times 13$, and $z = 4 \times 4 \times 5 \times 11$, and the roots of the three squares sought are $x = 3 \times 11 \times 13 = 429$, $y = 4 \times 5 \times 9 \times 13 = 2340$, and $z = 4 \times 4 \times 5 \times 11 = 880$: where it is evident that these are still less than those found above, from which we derive $x^2+y^2 = 3^2 \times 13^2 (121+3600) = 3^2 \times 13^2 \times 61^2$, $x^2+z^2 = 11^2 \times (1521+6400) = 11^2 \times 89^2$, $y^2+z^2 = 20^2 \times (13689+1936) = 20^2 \times 125^2$. 6. The last remark we shall make on this question is, that each answer easily furnishes a new one; for when we have * Thus, $(t^2-1)^2 = t^4 - 2t^2 + 1$, which multiplied by $p$ becomes $pt^4 - 2pt^2 + p$. Then adding, $4pt^2$ \rule{5.5em}{0.5pt} We have $pt^4+2pt^2+p = (t^2+1)^2 p$, as above.