Elements of Algebra

CHAP. V. OF ALGEBRA. 341 by $7n - 1$; that, in the same manner, the formula $7n + 5$ is the same as $7n - 2$, and $7n + 4$ the same as $7n - 3$. This being the case, it is evident, that the squares of the two classes of numbers, $7n + 1$, and $7n - 1$, if divided by $7$, will give the same remainder $1$; and that the squares of the two classes, $7n + 2$, and $7n - 2$, ought to resemble each other in the same respect, each leaving the remainder $4$. 77. In general, therefore, let the divisor be any number whatever, which we shall represent by the letter $d$, the different classes of numbers which result from it will be $dn$; $dn + 1, dn + 2, dn + 3, \&c.$ $dn - 1, dn - 2, dn - 3, \&c.$ in which the squares of $dn + 1$, and $dn - 1$, have this in common, that, when divided by $d$, they leave the remainder $1$, so that they belong to the same formula, $dn + 1$; in the same manner, the squares of the two classes $dn + 2$, and $dn - 2$, belong to the same formula $dn + 4$. So that we may conclude, generally, that the squares of the two kinds, $dn + a$, and $dn - a$, when divided by $d$, give a common remainder $a^2$, or that which remains in dividing $a^2$ by $d$. 78. These observations are sufficient to point out an infinite number of formulæ, such as $at^2 + bu^2$, which cannot by any means become squares. Thus, by considering the divisor $7$, it is easy to perceive, that none of these three formulæ, $7t^2 + 3u^2$, $7t^2 + 5u^2$, $7t^2 + 6u^2$, can ever become a square; because the division of $u^2$ by $7$ only gives the remainders $1, 2, \text{ or } 4$; and, in the first of these formulæ, there remains either $3, 6, \text{ or } 5$; in the second, $5, 3, \text{ or } 6$; and in the third, $6, 5, \text{ or } 3$; which cannot take place in square numbers. Whenever, therefore, we meet with such formulæ, we are certain that it is useless to attempt discovering any case, in which they can become squares: and, for this reason, the considerations, into which we have just entered, are of some importance. If, on the other hand, the formula proposed is not of this nature, we have seen in the last chapter, that it is sufficient to find a single case, in which it becomes a square, to enable us to deduce from it an infinite number of similar cases. The given formula, Art. 63, was properly $ax^2 + b$; and, as we usually obtain fractions for $x$, we supposed $x = \frac{t}{u}$, so that the problem, in reality, is to transform $at^2 + bu^2$ into a square.