Elements of Algebra

354 ELEMENTS PART II. $7n^2 + 1 = 4n^2 + 4np + p^2$, or $3n^2 = 4np + p^2 - 1$; which gives $n = \frac{2p + \sqrt{(7p^2 - 3)}}{3}$. At present, since $n > p$, and, consequently, greater than $p$, let us make $n = p + q$, and we shall have $p + 3q = \sqrt{(7p^2 - 3)}$; then, squaring both sides, $p^2 + 6pq + 9q^2 = 7p^2 - 3$, so that $6p^2 = 6pq + 9q^2 + 3$, or $2p^2 = 2pq + 3q^2 + 1$; whence we get $p = \frac{q + \sqrt{(7q^2 + 2)}}{2}$. Now, we have here $p > \frac{3q}{2}$; and, consequently, $p > q$; so that making $p = q + r$, we shall have $q + 2r = \sqrt{(7q^2 + 2)}$; the squares of which are $q^2 + 4qr + 4r^2 = 7q^2 + 2$; then $6q^2 = 4qr + 4r^2 - 2$, or $3q^2 = 2qr + 2r^2 - 1$; and, lastly, $q = \frac{r + \sqrt{(7r^2 - 3)}}{3}$. Since now $q > r$, let us suppose $q = r + s$, and we shall have $2r + 3s = \sqrt{(7r^2 - 3)}$; then $4r^2 + 12rs + 9s^2 = 7r^2 - 3$, or $3r^2 = 12rs + 9s^2 + 3$, or $r^2 = 4rs + 3s^2 + 1$, and $r = 2s + \sqrt{(7s^2 + 1)}$. Now, this formula is like the first; so that making $s = 0$, we shall obtain $r = 1$, $q = 1$, $p = 2$, and $n = 3$, or $m = 8$. But this calculation may be considerably abridged in the following manner, which may be adopted also in other cases. Since $7n^2 + 1 = m^2$, it follows that $m < 3n$. If, therefore, we suppose $m = 3n - p$, we shall have $7n^2 + 1 = 9n^2 - 6np + p^2$, or $2n^2 = 6np - p^2 + 1$; whence we obtain $n = \frac{3p + \sqrt{(7p^2 + 2)}}{2}$; so that $n < 3p$; for this reason we shall write $n = 3p - 2q$; and, squaring, we shall have $9p^2 - 12pq + 4q^2 = 7p^2 + 2$; or $2p^2 = 12pq - 4q^2 + 2$, and $p^2 = 6pq - 2q^2 + 1$; whence results $p = 3q + \sqrt{(7q^2 + 1)}$. Here, we can at once make $q = 0$, which gives $p = 1$, $n = 3$, and $m = 8$, as before. 103. Let $a = 8$, so that $8n^2 + 1 = m^2$, and $m < 3n$. Here, we must make $m = 3n - p$, and shall have $8n^2 + 1 = 9n^2 - 6np + p^2$, or $n^2 = 6np - p^2 + 1$; whence $n = 3p + \sqrt{(8p^2 + 1)}$, and this formula being al-