Elements of Algebra

326 ELEMENTS PART II. which $a$, or the first term, is a square. Let there be proposed, for example, the formula $f^2 + bx + cx^2$, which it is required to make a square. Here, let us suppose $\sqrt{(f^2 + bx + cx^2)} = f + \frac{mx}{n}$, and we shall have $f^2 + bx + cx^2 = f^2 + \frac{2fmx}{n} + \frac{m^2x^2}{n^2}$, in which equation the terms $f^2$ destroying each other, we may divide the remaining terms by $x$, so that $b + cx = \frac{2mf}{n} + \frac{m^2x}{n^2}$; or $n^2b + n^2cx = 2mnf + m^2x$, or $x(n^2c - m^2) = 2mnf - n^2b$; or, lastly, $x = \frac{2mnf - n^2b}{n^2c - m^2}$. If we now substitute this value instead of $x$, we have $\sqrt{(f^2 + bx + cx^2)} = f + \frac{2m^2f - mnb}{n^2c - m^2} = \frac{n^2cf + m^2f - mnb}{n^2c - m^2}$; and making $x = \frac{p}{q}$, we may, in the same manner as before, transform the expression $f^2q^2 + bpq + cp^2$, into a square, by making $p = 2mnf - n^2b$, and $q = n^2c - m^2$. 49. Here we have chiefly to distinguish the case in which $a=0$, that is to say, in which it is required to make a square of the formula $bx + cx^2$; for we have only to suppose $\sqrt{(bx + cx^2)} = \frac{mx}{n}$, from which we have the equation $bx + cx^2 = \frac{m^2x^2}{n^2}$; which, divided by $x$, and multiplied by $n^2$, gives $bn^2 + cn^2x = m^2x$; and, $x = \frac{bn^2}{m^2 - cn^2}$. If we seek, for example, all the triangular numbers that are at the same time squares, it will be necessary that $\frac{x^2+x}{2}$, which is the form of triangular numbers, must be a square; and, consequently, $2x^2 + 2x$ must also be a square. Let us, therefore, suppose $\frac{m^2x^2}{n^2}$ to be that square, and we shall have $2n^2x + 2n^2 = m^2x$, and $x = \frac{2n^2}{m^2 - 2n^2}$; in which value we may substitute, instead of $m$ and $n$, all pos-