Elements of Algebra

CHAP. XIV. OF ALGEBRA. 413 adopted at first, $p^2-2q^2$; for since $t = 3$, and $u = 2$, we have $r = 9$, and $s = 4$; wherefore $p = 81 - 32 = 49$, and $q = 72$; whence $p^2 - 2q^2 = 2401 - 10368 = -7967$. *** CHAP. XIV. Solution of some Questions that belong to this part of Algebra. 212. We have hitherto explained such artifices as occur in this part of Algebra, and such as are necessary for resolving any question belonging to it: it remains to make them still more clear, by adding here some of those questions with their solutions. 213. Question 1. To find such a number, that if we add unity to it, or subtract unity from it, we may obtain in both cases a square number. Let the number sought be $x$; then both $x + 1$, and $x - 1$ must be squares. Let us suppose for the first case $x + 1 = p^2$, we shall have $x = p^2 - 1$, and $x - 1 = p^2 - 2$, which must likewise be a square. Let its root, therefore, be represented by $p - q$; and we shall have $p^2 - 2 = p^2 - 2pq + q^2$; consequently, $p = \frac{q^2 + 2}{2q}$. Hence we obtain $x = \frac{q^4 + 4}{4q^2}$, in which we may give $q$ any value whatever, even a fractional one. If we therefore make $q = \frac{r}{s}$, so that $x = \frac{r^4 + 4s^4}{4r^2s^2}$, we shall have the following values for some small numbers: If $r = 1$, | $2$, | $1$, | $3$, | $4$, | and $s = 1$, | $1$, | $2$, | $1$, | $1$, | we have $x = \frac{5}{4}$, | $\frac{5}{4}$, | $\frac{65}{16}$, | $\frac{85}{36}$, | $\frac{65}{16}$, | 214. Question 2. To find such a number $x$, that if we add to it any two numbers, for example, 4 and 7, we obtain in both cases a square. According to this enunciation, the two formulæ, $x + 4$ and $x + 7$, must become squares. Let us therefore suppose the first $x + 4 = p^2$, which gives us $x = p^2 - 4$, and the