Elements of Algebra

CHAP. XIV. OF ALGEBRA. 417 On the contrary, when the number $a$ is the sum of two squares in fractional numbers, it is also the sum of two squares in integer numbers an infinite number of ways: and this we shall illustrate. 219. Question 5. To resolve, in as many ways as we please, a number, which is the sum of two squares, into another, that shall also be the sum of two squares. Let $f^2 + g^2$ be the given number, and let two other squares, $x^2$ and $y^2$, be required, whose sum $x^2 + y^2$ may be equal to the number $f^2 + g^2$. Here it is evident, that if $x$ is either greater or less than $f$, $y$, on the other hand, must be either less or greater than $g$: if, therefore, we make $x = f + pz$, and $y = g - qz$, we shall have $f^2 + 2fpz + p^2z^2 + g^2 - 2gqz + q^2z^2 = f^2 + g^2$, where the two terms $f^2$ and $g^2$ are destroyed; after which there remain only terms divisible by $z$. So that we shall have $2fp + p^2z - 2gq + q^2z = 0$, or $p^2z + q^2z = 2gq - 2fp$; therefore $z = \frac{2gq - 2fp}{p^2 + q^2}$, whence we get the following values for $x$ and $y$, namely, $x = \frac{2gpq + f(q^2 - p^2)}{p^2 + q^2}$, and \ldots $y = \frac{2fpq + g(p^2 - q^2)}{p^2 + q^2}$; in which we may substitute all possible numbers for $p$ and $q$. If 2, for example, be the number proposed, so that $f = 1$, and $g = 1$, we shall have $x^2 + y^2 = 2$; and because $x = \frac{2pq + q^2 - p^2}{p^2 + q^2}$, and $y = \frac{2pq + p^2 - q^2}{p^2 + q^2}$, if we make $p = 2$, and $q = 1$, we shall find $x = \frac{1}{5}$, and $y = \frac{7}{5}$. 220. Question 6. If $a$ be the sum of two squares, to find such a number, $x$, that $a + x$, and $a - x$, may become squares. Let $a = 13 = 9 + 4$, and let us make $13 + x = p^2$, and $13 - x = q^2$. Then we shall first have, by addition, $26 = p^2 + q^2$; and, by subtraction, $2x = p^2 - q^2$; consequently, the values of $p$ and $q$ must be such, that $p^2 + q^2$ may become equal to the number 26, which is also the sum of two squares, namely, of $25 + 1$. Now, since the question in reality is to resolve 26 into two squares, the greater of which may be expressed by $p^2$, and the less by $q^2$, we shall immediately have $p = 5$, and $q = 1$; so that $x = 12$. But we may resolve the number 26 into two squares in an EE