Elements of Algebra

CHAP. V. OF ALGEBRA. 339 second term $2u^2$, if divided by $4$, will leave the remainder $2$; so that the two terms together, when divided by $4$, leave a remainder of $2$, and therefore cannot form a square. Lastly, if we were to suppose $u$ an even number, as $2s$, and $t$ odd, so that $t^2$ is of the form $8n+1$, our formula would be changed into this, $24n + 3 + 8s^2$; which, divided by $8$, leaves $3$, and therefore cannot be a square. This demonstration extends to the formula $3t^2+(8n+2)u^2$; also to this, $(8m + 3)t^2 + 2u^2$, and even to this, $(8m+3)t^2 + (8n+2)u^2$; in which we may substitute for $m$ and $n$ all integer numbers, whether positive or negative. 72. But let us proceed farther, and consider the divisor $5$, with respect to which all numbers may be ranged under the five following classes: $5n, 5n + 1, 5n + 2, 5n + 3, 5n + 4$. We remark, in the first place, that if a number be of the first class, its square will have the form $25n^2$; and will consequently be divisible not only by $5$, but also by $25$. Every number of the second class will have a square of the form $25n^2 + 10n + 1$; and as dividing by $5$ gives the remainder $1$, this square will be contained in the formula $5n+1$. The numbers of the third class will have for their square $25n^2 + 20n + 4$; which, divided by $5$, gives $4$ for the remainder. The square of a number of the fourth class is $25n^2 + 30n + 9$; and if it be divided by $5$, there remains $4$. Lastly, the square of a number of the fifth class is $25n^2 + 40n + 16$; and if we divide this square by $5$, there will remain $1$. When a square number therefore cannot be divided by $5$, the remainder after division will always be $1$, or $4$, and never $2$, or $3$: hence it follows, that no square number can be contained in the formula $5n+2$, or $5n+3$. 73. From this it may be proved, that neither the formula $5t^2 + 2u^2$, nor $5t^2 + 3u^2$, can be a square. For, either $u$ is divisible by $5$, or it is not: in the first case, these formulae will be divisible by $5$, but not by $25$; therefore they cannot be squares. On the other hand, if $u$ be not divisible by $5$, $u^2$ will either be of the form $5n+1$, or $5n+4$. In the first of these cases, the formula $5t^2 + 2u^2$ becomes $5t^2 + 10n + 2$; which, divided by $5$, leaves a remainder of $2$; and the formula $5t^2 + 3u^2$ becomes $5t^2 + 15n + 3$; which, being divided by $5$, gives a remainder of $3$; so that neither the one nor the other can be a square. With regard to the case of $u^2 = 5n+4$, the first formula becomes $5t^2 + 10n + 8$; Z 2