Elements of Algebra

CHAP. VI. OF ALGEBRA. 347 It is here required to have $7x^2 + 2 = y^2$, or $a = 7$, and $b=2$; and the known case immediately occurs, that is to say, $x=1$; so that $x=f=1$, and $y=g=3$. If we next consider the equation $7n^2+1=m^2$, we easily find also that $n=3$, and $m=8$; whence $x=8f+3g$, and $y=8g+21f$. We shall therefore have the following results: $x=f=1 \mid 17 \mid 271$ $y=g=3 \mid 45 \mid 717, \ldots$ 91. Question 5. To find all the triangular numbers, that are at the same time pentagons. Let the root of the triangle be $p$, and that of the pentagon $q$: then we must have $\frac{p^2+p}{2} = \frac{3q^2-q}{2}$, or $3q^2-q=p^2+p$; and, in endeavouring to find $q$, we shall first have $q^2 = \frac{1}{3}q + \frac{p^2+p}{3}$, and $q = \frac{1}{6} \pm \sqrt{(\frac{1}{36} + \frac{p^2+p}{3})}$, or $q = \frac{1 \pm \sqrt{(12p^2+12p+1)}}{6}$. Consequently, it is required to make $12p^2+12p+1$ become a square, and that in integer numbers. Now, as there is here a middle term $12p$, we shall begin with making $p = \frac{x-1}{2}$, by which means we shall have $12p^2=3x^2-6x+3$, and $12p=6x-6$; consequently, $12p^2+12p+1=3x^2-2$; and it is this last quantity, which at present we are required to transform into a square. If, therefore, we make $3x^2-2=y^2$, we shall have $p = \frac{x-1}{2}$, and $q = \frac{1+y}{6}$; so that all depends on the formula $3x^2-2=y^2$; and here we have $a=3$, and $b=-2$. Farther, we have a known case, $x=f=1$, and $y=g=1$; lastly, in the equation $m^2 = 3n^2+1$, we have $n=1$, and $m=2$; therefore we find the following values both for $x$ and $y$, and for $p$ and $q$: First, $x=2f+g$, and $y=2g+3f$; then, $x=f=1 \mid 3 \mid 11 \mid 41$ $y=g=1 \mid 5 \mid 19 \mid 71$ $p=0 \mid 1 \mid 5 \mid 20$ $q=\frac{1}{3} \mid 1 \mid \frac{10}{3} \mid 12$ or $q=0 \mid -\frac{2}{3} \mid -3 \mid -\frac{35}{3}$ because we have also $q = \frac{1-y}{6}$.