Elements of Algebra

340 ELEMENTS PART II. which, divided by 5, leaves 3; and the other becomes $5t^2 + 15n + 12$, which, divided by 5, leaves 2; so that in this case also, neither of the two formulæ can be a square. For a similar reason, we may remark, that neither the formula $5t^2 + (5n + 2)u^2$, nor $5t^2 + (5n + 3)u^2$, can become a square, since they leave the same remainders that we have just found. We might even in the first term write $5mt^2$, instead of $5t^2$, provided $m$ be not divisible by 5. 74. Since all the even squares are contained in the formula $4n$, and all the odd squares in the formula $4n + 1$; and, consequently, since neither $4n + 2$, nor $4n + 3$, can become a square, it follows that the general formula $(4m + 3)t^2 + (4n + 3)u^2$ can never be a square. For if $t$ be even, $t^2$ will be divisible by 4, and the other term, being divided by 4, will give 3 for a remainder; and, if we suppose the two numbers $t$ and $u$ odd, the remainders of $t^2$ and of $u^2$ will be 1; consequently, the remainder of the whole formula will be 2: now, there is no square number, which, when divided by 4, leaves a remainder of 2. We shall remark, also, that both $m$ and $n$ may be taken negatively, or $= 0$, and still the formulæ $3t^2 + 3u^2$, and $3t^2 - u^2$, cannot be transformed into squares. 75. In the same manner as we have found for a few divisors, that some kinds of numbers can never become squares, we might determine similar kinds of numbers for all other divisors. If we take the divisor 7, we shall have to distinguish seven different kinds of numbers, the squares of which we shall also examine. Kinds of numbers. Their squares are of the kind. 1. $7n$ $49n^2$ $7n$ 2. $7n + 1$ $49n^2 + 14n + 1$ $7n + 1$ 3. $7n + 2$ $49n^2 + 28n + 4$ $7n + 4$ 4. $7n + 3$ $49n^2 + 42n + 9$ $7n + 2$ 5. $7n + 4$ $49n^2 + 56n + 16$ $7n + 2$ 6. $7n + 5$ $49n^2 + 70n + 25$ $7n + 4$ 7. $7n + 6$ $49n^2 + 84n + 36$ $7n + 1$. Therefore, since the squares which are not divisible by 7, are all contained in the three formulæ $7n + 1$, $7n + 2$, $7n + 4$, it is evident, that the three other formulæ, $7n + 3$, $7n + 5$, and $7n + 6$, do not agree with the nature of squares. 76. To make this conclusion still more apparent, we shall remark, that the last kind, $7n + 6$, may be also expressed