Elements of Algebra

416 ELEMENTS PART II. by demanding such numbers, $x$, for any number, $a$, that both $a + x$, and $a - x$ may be squares, the solution would frequently become impossible; namely, in all cases in which $a$ was not the sum of two squares. Now, we have already seen, that, between 1 and 50, there are only the following numbers that are the sums of two squares, or that are contained in the formula $x^2 + y^2$: 1, 2, 4, 5, 8, 9, 10, 13, 16, 17, 18, 20, 25, 26, 29, 32, 34, 36, 37, 40, 41, 45, 49, 50. So that the other numbers, comprised between 1 and 50, which are, 3, 6, 7, 11, 12, 14, 15, 19, 21, 22, 23, 24, 27, 28, 30, 31, 33, 35, 38, 39, 42, 43, 44, 46, 47, 48, cannot be resolved into two squares; consequently, whenever $a$ is one of these last numbers, the question will be impossible; which may be thus demonstrated: Let $a + x = p^2$, and $a - x = q^2$, then the addition of the two formulæ will give $2a = p^2 + q^2$; therefore $2a$ must be the sum of two squares. Now, if $2a$ be such a sum, $a$ will be so likewise*; consequently, when $a$ is not the sum of two squares, it will always be impossible for $a + x$, and $a - x$, to be each squares at the same time. 218. As 3 is not the sum of two squares, it follows, from what has been said, that, if $a = 3$, the question is impossible. It might, however, be objected, that there are, perhaps, two fractional squares whose sum is 3; but we answer that this also is impossible: for if $\frac{p^2}{q^2} + \frac{r^2}{s^2} = 3$, and if we were to multiply by $q^2s^2$, we should have $3q^2s^2 = p^2s^2 + q^2r^2$; and the second side of this equation, which is the sum of two squares, would be divisible by 3; but we have already seen (Art. 170) that the sum of two squares, that are prime to each other, can have no divisors, except numbers, which are themselves sums of two squares. The numbers 9 and 45, it is true, are divisible by 3, but they are also divisible by 9, and even each of the two squares that compose both the one and the other, is divisible by 9, since $9 = 3^2 + 0^2$, and $45 = 6^2 + 3^2$; which is therefore a different case, and does not enter into consideration here. We may rest assured, therefore, of this conclusion; that if a number, $a$, be not the sum of two squares in integer numbers, it will not be so in fractions. * For, let $x^2 + y^2 = 2a$; and put $x = s + d$, and $y = s - d$; then $(s+d)^2 + (s-d)^2 = 2s^2 + 2d^2$: that is, $x^2 + y^2 = 2s^2 + 2d^2 = 2a$, or $s^2 + d^2 = a$.