Elements of Algebra

CHAP. VII. OF ALGEBRA. 357 $z = 0$, $u = v + x = 3$, $q = r + s = 71$, $y = 1$, $t = u + v = 5$, $p = q + r = 109$, $x = y + z = 1$, $s = 6t + u = 33$, $n = p + q = 180$, $v = x + y = 2$, $r = s + t = 38$, $m = 3n + p = 649$. So that 180 is the least number, after 0, which we can substitute for $n$, in order that $13n^2 + 1$ may become a square. 106. This example sufficiently shews how prolix these calculations may be in particular cases; and when the numbers in question are greater, we are often obliged to go through ten times as many operations as we had to perform for the number 13. As we cannot foresee the numbers that will require such tedious calculations, we may with propriety avail ourselves of the trouble which others have taken; and, for this purpose, a Table is subjoined to the present chapter, in which the values of $m$ and $n$ are calculated for all numbers, $a$, between 2 and 100; so that in the cases which present themselves, we may take from it the values of $m$ and $n$, which answer to the given number $a$. 107. It is proper, however, to remark, that, for certain numbers, the letters $m$ and $n$ may be determined generally. This is the case when $a$ is greater, or less than a square, by 1 or 2; it will be proper, therefore, to enter into a particular analysis of these cases. 108. In order to this, let $a = e^2 - 2$; and since we must have $(e^2 - 2)n^2 + 1 = m^2$, it is clear that $m < en$; therefore we shall make $m = en - p$, from which we have $(e^2 - 2)n^2 + 1 = e^2n^2 - 2enp + p^2$, or $2n^2 = 2enp - p^2 + 1$; therefore $n = \frac{ep + \sqrt{(e^2p^2 - 2p^2 + 2)}}{2}$; and it is evident that if we make $p = 1$, this quantity becomes rational, and we have $n = e$, and $m = e^2 - 1$. For example, let $a = 23$, so that $e = 5$; we shall then have $23n^2 + 1 = m^2$, if $n = 5$, and $m = 24$. The reason of which is evident from another consideration; for if, in the case of $a = e^2 - 2$, we make $n = e$, we shall have $an^2 + 1 = e^4 - 2e^2 + 1$; which is the square of $e^2 - 1$. 109. Let $a = e^2 - 1$, or less than a square by unity. First, we must have $(e^2 - 1)n^2 + 1 = m^2$; then, because, as before, $m < en$, we shall make $m = en - p$; and this being done, we have $(e^2 - 1)n^2 + 1 = e^2n^2 - 2enp + p^2$, or $n^2 = 2enp - p^2 + 1$;