Elements of Algebra

CHAP. XIV. OF ALGEBRA. 447 found three values, $x = a$, $y = b$, and $z = c$, so that $a^2 + b^2 = \square$, $a^2 + c^2 = \square$, and $b^2 + c^2 = \square$, the three following values will likewise be satisfactory, namely, $x=ab$, $y=bc$, and $z=ac$. Then we must have $x^2 + y^2 = a^2b^2 + b^2c^2 = b^2(a^2 + c^2) = \square$, $x^2 + z^2 = a^2b^2 + a^2c^2 = a^2(b^2 + c^2) = \square$, $y^2 + z^2 = a^2c^2 + b^2c^2 = c^2(a^2 + b^2) = \square$. Now, as we have just found $x = a = 3 \times 11 \times 13$, $y = b = 4 \times 5 \times 9 \times 13$, and $z = c = 4 \times 4 \times 5 \times 11$, we have, therefore, according to the new solution, $x = ab = 3 \times 4 \times 5 \times 9 \times 11 \times 13 \times 13$, $y = bc = 4 \times 4 \times 4 \times 5 \times 5 \times 9 \times 11 \times 13$, $z = ac = 3 \times 4 \times 4 \times 5 \times 11 \times 11 \times 13$. And all these three values being divisible by $3 \times 4 \times 5 \times 11 \times 13$, are reducible to the following, $x = 9 \times 13$, $y = 3 \times 4 \times 4 \times 5$, and $z = 4 \times 11$; or $x=117$, $y=240$, and $z=44$, which are still less than those which the preceding solution gave, and from them we deduce $x^2 + y^2 = 71289 = 267^2$, $x^2 + z^2 = 15625 = 125^2$, $y^2 + z^2 = 59536 = 244^2$. 239. Question 16. Required two such numbers, $x$ and $y$, that each being added to the square of the other, may make a square; that is, that $x^2 + y = \square$, and $y^2 + x = \square$. If we begin with supposing $x^2 + y = p^2$, and from that deduce $y = p^2 - x^2$, we shall have for the other formula $p^4 - 2p^2x^2 + x^4 + x = \square$, which it would be difficult to resolve. Let us, therefore, suppose one of the formulae $x^2 + y = (p-x)^2 = p^2 - 2px + x^2$; and, at the same time, the other $y^2 + x = (q-y)^2 = q^2 - 2qy + y^2$, and we shall thus obtain the two following equations, $y + 2px = p^2$, and $x + 2py = q^2$, from which we easily deduce $x = \frac{2qp^2-q^2}{4pq-1}$, and $y = \frac{2pq^2-p^2}{4pq-1}$, in which $p$ and $q$ are indeterminate. Let us, therefore, suppose, for example, $p=2$, and $q=3$, then we shall have