Elements of Algebra

CHAP. VI. OF ALGEBRA. 345 pose we need only make $x=f$, and $y=g$; but it is also evident, that this supposition would not furnish a new case in addition to that already known. We shall, therefore, suppose that we have already found such a number for $n$, that $an^2 + 1$ is a square, or that $an^2 + 1 = m^2$; which being laid down, we have $m^2 - an^2 = 1$; and multiplying by this equation the one we had last, we find also $y^2 - ax^2 = (g^2 - af^2) \times (m^2 - an^2) = g^2m^2 - af^2m^2 - ag^2n^2 + a^2f^2n^2$. Let us now suppose $y = gm + afn$, and we shall have $g^2m^2 + 2afgmn + a^2f^2n^2 - ax^2 = g^2m^2 - af^2m^2 - ag^2n^2 + a^2f^2n^2$, in which the terms $g^2m^2$ and $a^2f^2n^2$ are destroyed; so that there remains $ax^2 = af^2m^2 + ag^2n^2 + 2afgmn$, or $x^2 = f^2m^2 + 2fgmn + g^2n^2$. Now, this formula is evidently a square, and gives $x = fm + gn$. Hence we have obtained the same formulæ for $x$ and $y$ as before. 87. It will be necessary to render this solution more evident, by applying it to some examples. *Question* 1. To find all the integer values of $x$, that will make $2x^2 - 1$, a square, or give $2x^2 - 1 = y^2$. Here, we have $a = 2$ and $b = -1$; and a satisfactory case immediately presents itself, namely, that in which $x=1$, and $y=1$: which gives us $f=1$, and $g=1$. Now, it is farther required to determine such a value of $n$, as will give $2n^2 + 1 = m^2$; and we see immediately, that this obtains when $n=2$, and consequently $m=3$; so that every case, which is known for $f$ and $g$, giving us these new cases $x = 3f + 2g$, and $y = 3g + 4f$, we derive from the first solution, ($f=1$, and $g=1$,) the following new solutions: If $\begin{cases} f=1, \\ g=1, \end{cases}$ Then $\begin{cases} x = 5, & 29, & 169, & \text{\&c.} \\ y = 7, & 41, & 239, & \text{\&c.} \end{cases}$ 88. *Question* 2. To find all the triangular numbers, that are at the same time squares. Let $z$ be the triangular root; then $\frac{z^2+z}{2}$ is the triangle, which is to be also a square; and if we call $x$ the root of this square, we have $\frac{z^2+z}{2} = x^2$: multiplying by 8, we have $4z^2 + 4z = 8x^2$; and also adding 1 to each side, we have $4z^2 + 4z + 1 = (2z + 1)^2 = 8x^2 + 1$. Hence the question is to make $8x^2 + 1$ become a square;