Elements of Algebra

346 ELEMENTS PART II. for, if we find $8x^2 + 1 = y^2$, we shall have $y = 2z + 1$, and, consequently, the triangular root required will be $z = \frac{y-1}{2}$. Now, we have $a = 8$, and $b = 1$, and a satisfactory case immediately occurs, namely, $f = 0$ and $g = 1$. It is farther evident, that $8n^2 + 1 = m^2$, if we make $n = 1$, and $m = 3$; therefore $x = 3f + g$, and $y = 3g + 8f$; and since $z = \frac{y-1}{2}$, we shall have the following solutions: $x = f = 0$ | 1 | 6 | 35 | 204 | 1189 $y = g = 1$ | 3 | 17 | 99 | 577 | 3363 $z = \frac{y-1}{2} = 0$ | 1 | 8 | 49 | 288 | 1681, &c. 89. Question 3. To find all the pentagonal numbers, which are at the same time squares. If the root be $z$, the pentagon will be $\frac{3z^2 - z}{2}$, which we shall make equal to $x^2$, so that $3z^2 - z = 2x^2$; then multiplying by 12, and adding unity, we have $36z^2 - 12z + 1 = (6z - 1)^2 = 24x^2 + 1$; also, making $24x^2 + 1 = y^2$, we have $y = 6z - 1$, and $z = \frac{y+1}{6}$. Since $a = 24$, and $b = 1$, we know the case $f = 0$, and $g = 1$; and as we must have $24n^2 + 1 = m^2$, we shall make $n = 1$, which gives $m = 5$; so that we shall have $x = 5f + g$ and $y = 5g + 24f$; and not only $z = \frac{y+1}{6}$, but also $z = \frac{1-y}{6}$, because we may write $y = 1 - 6z$; whence we find the following results: $x = f = 0$ | 1 | 10 | 99 | 980 $y = g = 1$ | 5 | 49 | 485 | 4801 $z = \frac{y+1}{6} = \frac{1}{3}$ | 1 | $\frac{25}{3}$ | 81 | $\frac{2401}{3}$ or $z = \frac{1-y}{6} = 0$ | $-\frac{2}{3}$ | $-8$ | $-\frac{242}{3}$ | $-800$, &c. 90. Question 4. To find all the integer square numbers, which, if multiplied by 7 and increased by 2, become squares.