Elements of Algebra

CHAP. VII. ADDITIONS. 573 The preceding problem may be resolved more easily by mere trial; for when we have arrived at the equation, $x^2 = 1171 - 7y^2$, we shall only have to try, for $y$, all the whole numbers, whose squares multiplied by $7$ do not exceed $1171$; that is to say, all the numbers $\le \sqrt{\frac{1171}{7}}$, or $\le 13$. It is the same with all the equations, in which A is a negative number; for when we are brought to the equation, $x^2 = B + Ay^2$, where making $A = -a$, and $x^2 = B - ay^2$, it is evident, that the satisfactory values of $y$, if there are any, can only be found among the numbers, $\le \sqrt{\frac{B}{a}}$. So that I have not given particular methods for the case of A negative, only because these methods are intimately connected with those concerning the case of A positive, and because all these methods, being so nearly alike, reciprocally illustrate and confirm each other. 82. *Example 2.* Let us now give some examples for the case of A positive, and let it be proposed to find all the whole numbers, which we may take for $y$, in order that the radical quantity, $\sqrt{(13y^2 + 101)}$, may become rational. Here, we shall have (Art. 64), $A = 13$, $B = 101$; and the equation to be resolved in integers will be, $x^2 - 13y^2 = 101$, in which, because $101$ is not divisible by any square, $y$ must be prime to $101$. We shall therefore make (Art. 65), $x = ny - 101z$, and $n^2 - 13$ must be divisible by $101$, taking $n < \frac{101}{2}$, $< 51$. I find $n = 35$, which gives $n^2 = 1225$, and $n^2 - 13 = 1212 = 101 \times 12$; so that we may take $n = \pm 35$, and substituting $\pm 35y - 101z$, instead of $x$, we shall have an equation wholly divisible by $101$, which, after the division, will be $12y^2 \mp 70yz + 101z^2 = 1$. In order to resolve this equation, let us also employ the method of Article 70. Let us make $D' = 12$, $D = 101$, $n = \pm 35$; but, instead of the letter $t$, we shall preserve the letter $y$, and shall only change $z$ into $y'$, as in the preceding example. 1st. If $n = 35$, we shall make the following calculation: $m = \frac{35}{12} = 3$, $n' = 35 - 3 \times 12 = -1$, $n'' = \frac{1 - 13}{12} = -1$, $y = 3y' + y''$,