Elements of Algebra

334 ELEMENTS PART II. If $m=2$, and $n=1$, we have $x=-6$, and, consequently, $5x^2+3x+7=169=13^2$. But if $m=-2$, and $n=1$, we find $x=18$, and $5x^2+3x+7=1681=41^2$. 61. Let us now consider the formula, $7x^2+15x+13$, in which we must begin with the supposition of $x=\frac{t}{u}$. Having substituted and multiplied $u^2$, we obtain $7t^2+15tu+13u^2$, which must be a square. Let us therefore try to adopt some small numbers as the values of $t$ and $u$. If $t=1$, and $u=1$, $\}$ the formula will become $\left\{\begin{array}{l} =35 \\ =71 \\ =11 \\ =121. \end{array}\right.$ $t=2$, and $u=1$, $t=2$, and $u=-1$, $t=3$, and $u=1$, Now, 121 being a square, it is proof that the value of $x=3$ answers the required condition; let us therefore suppose $x=y+3$, and we shall have, by substituting this value in the formula, $7y^2+42y+63+15y+45+13$, or $7y^2+57y+121$. Therefore let the root be represented by $11+\frac{my}{n}$, and we shall have $7y^2+57y+121 = 121 + \frac{22my}{n} + \frac{m^2y^2}{n^2}$, or $7n^2y+57n^2=22mn+m^2y$; whence $y=\frac{57n^2-22mn}{m^2-7n^2}$, and $x=\frac{36n^2-22mn+3m^2}{m^2-7n^2}$. Suppose, for example, $m=3$, and $n=1$; we shall then find $x=-\frac{3}{2}$, and the formula becomes $7x^2+15x+13=\frac{25}{4}=(\frac{5}{2})^2$. If $m=1$, and $n=1$, we find $x=-\frac{17}{6}$; if $m=3$, and $n=-1$, we have $x=\frac{129}{2}$, and the formula $7x^2+15x+13=\frac{120409}{4}=(\frac{347}{2})^2$. 62. But frequently it is only lost labour to endeavour to find a case, in which the proposed formula may become a square. We have already said that $3x^2+2$ is one of those unmanageable formulæ; and, by giving it, according to this rule, the form $3t^2+2u^2$, we shall perceive that, whatever values we give to $t$ and $u$, this quantity never becomes a square number. As the formulæ of this kind are very