Elements of Algebra

CHAP. VII. ADDITIONS. 567 $r = \frac{\alpha t + \beta u + \gamma}{\delta}, s = \frac{\alpha' t + \beta' u + \gamma'}{\delta'}$, $\alpha, \beta, \gamma, \delta, \alpha', \beta', \gamma', \delta'$, being known whole numbers, and $t$, $u$, being given by the formulæ of Art. 75, in which the exponent $m$ may be any positive whole number; thus, the question is reduced to finding what value we must give to $m$, in order that the values of $r$ and $s$ may be whole numbers. 78. I observe, first, that it is always possible to find a value of $u$ divisible by any given number, $\Delta$; for, supposing $u = \Delta \omega$, the equation, $t^2 - A u^2 = 1$, will become $t^2 - A \Delta^2 \omega^2 = 1$, which is always resolvible in whole numbers; and we shall find the least values of $t$ and $\omega$, by making the same calculation as before, only taking $A \Delta^2$, instead of $A$. Now, as these values also satisfy the equation $t^2 - A u^2 = 1$, they will necessarily be contained in the formulæ of Art. 75. Thus, we shall necessarily have a value of $m$, which will make the expression of $u$ divisible by $\Delta$. Let us denote this value of $m$ by $\mu$, and I say that, if we make $m = 2\mu$, in the general expressions of $t$ and $u$ of the Article just quoted, the value of $u$ will be divisible by $\Delta$; and that of $t$ being divided by $\Delta$ will give 1 for a remainder. For, if we express by $t'$ and $v'$ the values of $t$ and $u$, in which $m = \mu$, and by $T''$ and $v''$ those in which $m = 2\mu$, we shall have (Art. 75), $t' \pm v' \sqrt{A} = (T \pm v \sqrt{A})^\mu$, and $T'' \pm v'' \sqrt{A} = (T \pm v \sqrt{A})^{2\mu}$; therefore, $(t' \pm v' \sqrt{A})^2 = (T'' \pm v'' \sqrt{A})$, that is to say, comparing the rational part of the first side with the rational part of the second, and the irrational with the irrational, $T'' = t'^2 + A v'^2$, and $v'' = 2t'v'$; hence, since $v'$ is divisible by $\Delta$, $v''$ will be so likewise; and $T''$ will leave the same remainder that $t'^2$ would leave; but we have $t'^2 - A v'^2 = 1$ (hyp.), therefore $t'^2 - 1$ must be divisible by $\Delta$, and even by $\Delta^2$, since $v'^2$ is so already; wherefore, $t'^2$, and, consequently, $T''$ likewise, being divided by $\Delta$, will leave the remainder 1.