Elements of Algebra

408 ELEMENTS PART II. the one even and the other odd: and as in both cases the dif- ference of two squares may become a square, we must con- sider these two cases separately. 2. Let us, therefore, first begin by supposing both the numbers $x$ and $y$ odd, and that $x = p + q$, and $y = p - q$; then one of the two numbers $p$ and $q$ must necessarily be even, and the other odd. We have also $x^2 - y^2 = 4pq$, and $x^2 + y^2 = 2p^2 + 2q^2$; therefore our formula $x^4 - y^4 = 4pq(2p^2 + 2q^2)$; and as this must be a square, its fourth part, $pq(2p^2 + 2q^2) = 2pq(p^2 + q^2)$, must also be a square. Also, since the factors of this formula have no common di- visor (because if $p$ is even, $q$ must be odd), each of these fac- tors, $2p$, $q$, and $p^2 + q^2$, must be a square. In order, there- fore, that the first two may become squares, let us suppose $2p = 4r^2$, or $p = 2r^2$, and $q = s^2$; in which $s$ must be odd, and the third factor, $4r^4 + s^4$, must likewise be a square. 3. Now, since $s^4 + 4r^4$ is the sum of two squares, the first of which, $s^4$, is odd, and the other, $4r^4$, is even, let us make the root of the first $s^2 = t^2 - u^2$, in which let $t$ be odd, and $u$ even; and the root of the second, $2r^2 = 2tu$, or $r^2 = tu$, where $t$ and $u$ are prime to each other. 4. Since $tu = r^2$ must be a square, both $t$ and $u$ must be squares also. If, therefore, we suppose $t = m^2$, and $u = n^2$, (representing an odd number by $m$, and an even number by $n$), we shall have $s^2 = m^4 - n^4$; so that here also, it is re- quired to make the difference of two biquadrates, namely, $m^4 - n^4$, a square. Now, it is obvious, that these numbers would be much less than $x$ and $y$, since they are less than $r$ and $s$, which are themselves evidently less than $x$ and $y$. If a solution, therefore, were possible in great numbers, and $x^4 - y^4$ were a square, there must also be one possible for numbers much less; and this last would lead us to another solution for numbers still less, and so on. 5. Now, the least numbers for which such a square can be found, are in the case where one of the biquadrates is $0$, or where it is equal to the other biquadrate. In the first case, we must have $n = 0$; therefore $u = 0$, and also $r = 0$, $p = 0$, and, lastly, $x^4 - y^4 = 0$, or $x^4 = y^4$; which is a case that does not belong to the present question; if $n = m$, we shall find $t = u$, then $s = 0$, $q = 0$, and, lastly, also $x = y$, which does not here enter into consideration. 207. It might be objected, that since $m$ is odd, and $n$ even, the last difference is no longer similar to the first; and that, therefore, we can form no analogous conclusions from it with respect to smaller numbers. But it is sufficient that the first difference has led us to the second; and we shall