Elements of Algebra

CHAP. XIII. OF ALGEBRA. 407 $p² - q² = x²$, we must have, as we have already seen, $p = r² + s²$, and $q = 2rs$; for then $x² = (r² - s²)²$, and $x = r² - s²$. 6. Now, $y²$ must likewise be a square; and since we had $y² = 2pq$, we shall now have $y² = 4rs(r² + s²)$; so that this formula must be a square; therefore $rs(r² + s²)$ must also be a square: and let it be observed, that $r$ and $s$ are numbers prime to each other; so that the three factors of this formula, namely, $r$, $s$, and $r² + s²$, have no common divisor. 7. Again, when a product of several factors, that have no common divisor, must be a square, each factor must itself be a square; so that making $r = t²$, and $s = u²$, we must have $t⁴ + u⁴ = \Box$. If, therefore, $x⁴ + y⁴$ were a $\Box$, our formula $t⁴ + u⁴$, which is, in like manner, the sum of two biquadrates, would also be a $\Box$. And it is proper to observe here, that since $x² = t⁴ - u⁴$, and $y² = 4t²u²(t⁴ + u⁴)$ the numbers $t$ and $u$ will evidently be much smaller than $x$ and $y$, since $x$ and $y$ are even determined by the fourth powers of $t$ and $u$, and must therefore become much greater than these numbers. 8. It follows, therefore, that if we could assign, in numbers however great, two biquadrates, such as $x⁴$ and $y⁴$, whose sum might be a square, we could deduce from it a number, formed by the sum of two much less biquadrates, which would also be a square; and this new sum would enable us to find another of the same nature, still less, and so on, till we arrived at very small numbers. Now, such a sum not being possible in very small numbers, it evidently follows, that there is not one which we can express by very great numbers. 9. It might indeed be objected, that such a sum does exist in very small numbers; namely, in the case which we have mentioned, when one of the two biquadrates becomes nothing: but we answer, that we shall never arrive at this case, by coming back from very great numbers to the least, according to the method which has been explained; for if in the small sum, or the reduced sum, $t⁴ - u⁴$, we had $t = 0$, or $u = 0$, we should necessarily have $y² = 0$ in the great sum; but this is a case which does not here enter into consideration. 206. Let us proceed to the second proposition, and prove also that the difference of two biquadrates, or $x⁴ - y⁴$, can never become a square, except in the cases of $y = 0$, and $y = x$. 1. We may consider the numbers $x$ and $y$ as prime to each other, and consequently, as being either both odd, or