Elements of Algebra

430 ELEMENTS PART II. *Demonstration*. We have seen that $p$ is odd, and $q$ even, because $p^2 + q^2$ cannot be a square, except when $q = 2rs$, and $p = r^2 - s^2$; and $p^2 + 3q^2$ cannot be a square, except when $q = 2tu$, and $p = t^2 - 3u^2$, or $p = 3u^2 - t^2$. Now, as in both cases $q$ must be a double product, let us suppose for both, $q = 2abcd$; and, for the first formula, let us make $r = ab$, and $s = cd$; for the second, let $t = ac$, and $u = bd$. We shall have for the former $p = a^2b^2 - c^2d^2$, and for the latter $p = a^2c^2 - 3b^2d^2$, or $p = 3b^2d^2 - a^2c^2$, and these two values must be equal; so that we have either $a^2b^2 - c^2d^2 = a^2c^2 - 3b^2d^2$, or $a^2b^2 - c^2d^2 = 3b^2d^2 - a^2c^2$; and it will be perceived that the numbers $a, b, c$, and $d$, are each less than $p$ and $q$. We must, however, consider each case separately: the first gives $a^2b^2 + 3b^2d^2 = c^2d^2 + a^2c^2$, or $b^2(a^2 + 3d^2) = c^2(a^2 + d^2)$, whence $\frac{b^2}{c^2} = \frac{a^2+d^2}{a^2+3d^2}$ a fraction that must be a square. Now, the numerator and denominator can here have no other common divisor than 2, because their difference is $2d^2$. If, therefore, 2 were a common divisor, both $\frac{a^2+d^2}{2}$, and $\frac{a^2+3d^2}{2}$, must be a square; but the numbers $a$ and $d$ are in this case both odd, so that their squares have the form $8n+1$, and the formula $\frac{a^2+3d^2}{2}$ is contained in the expression $4n+2$, and cannot be a square; wherefore 2 cannot be a common divisor; the numerator $a^2+d^2$, and the denominator $a^2+3d^2$ are therefore prime to each other, and each of them must of itself be a square. But these formulae are similar to the former, and if the last were squares, similar formulae, though composed of the smallest numbers, would have also been squares; so that we conclude, reciprocally, from our not having found squares in small numbers, that there are none in great. This conclusion, however, is not admissible, unless the second case, $a^2b^2 - c^2d^2 = 3b^2d^2 - a^2c^2$, furnishes a similar one. Now, this equation gives $a^2b^2 + a^2c^2 = 3b^2d^2 + c^2d^2$, or $a^2(b^2+c^2) = d^2(3b^2+c^2)$; and, consequently, $\frac{a^2}{d^2} = \frac{b^2+c^2}{3b^2+c^2} = \frac{c^2+b^2}{c^2+3b^2}$; so that as this fraction ought to be a square, the foregoing conclusion is fully confirmed; for, if in great numbers there were cases in which $p^2+q^2$, and $p^2+3q^2$, were squares, such cases must have also existed with regard to smaller numbers; but this is not the fact.