Elements of Algebra

CHAP. XIV. OF ALGEBRA. 439 $x = 434657$ $y = 420968$ $z = 150568$ we have $x + y = 855625 = (925)^2$ $x + z = 585225 = (765)^2$ and $y + z = 571536 = (756)^2$ also, $x - y = 13689 = (117)^2$ $x - z = 284089 = (533)^2$ and $y - z = 270400 = (520)^2$. The Table which has been given, would enable us to find other numbers also, by supposing $f^2 = 9$, and $k^2 = 4$, $g^2 = 121$, and $h^2 = 4$; for then $t^2 = 13 \times 5 \times 5 \times 13 \times 9 \times 25 = 9 \times 25 \times 25 \times 169$, and $t = 3 \times 5 \times 5 \times 13 = 975$. Now, as $f = 3$, $g = 11$, $k = 2$, and $h = 2$, we have $a = fh = 6$, and $b = gk = 22$; consequently, $p = a^2 - b^2 = -448$, $q = a^2 + b^2 = 520$, and $r = 2ab = 264$; whence $2x = t^2 + p^2 + q^2 = 950625 + 200704 + 270400 = 1421729$, and $x = \frac{1421729}{2}$; wherefore $y = x - p^2 = \frac{1020321}{2}$, and $z = x - q^2 = \frac{880929}{2}$. Now, it is to be observed, that if these numbers have the property required, they will preserve it by whatever square they are multiplied. If, therefore, we take them four times greater, the following numbers must be equally satisfactory: $x = 2843458$, $y = 2040642$, and $z = 1761858$; and as these numbers are greater than the former, we may consider the former as the least that the question admits of. 236. Question 14. Required three such squares, that the difference of every two of them may be a square. The preceding solution will serve to resolve the present question. In fact, if $x$, $y$, and $z$, are such numbers that the following formulæ, namely, $x + y = \square$, $x - y = \square$, $x + z = \square$, $x - z = \square$, $y + z = \square$, $y - z = \square$, may become squares; it is evident, likewise, that the product $x^2 - y^2$ of the first and second, the product $x^2 - z^2$ of the third and fourth, and the product $y^2 - z^2$ of the fifth and sixth, will be squares; and, consequently, $x^2$, $y^2$, and $z^2$, will be three such squares as are sought. But these numbers would be very great, and there are, doubtless, less numbers that will satisfy the question; since, in order that $x^2 - y^2$ may become a square, it is not necessary that $x+y$, and $x-y$, should be squares: for example, $25-9$ is a square,