Elements of Algebra

CHAP. XIV. OF ALGEBRA. 429 it should be observed, that $s$ and $r$ must be odd numbers, and also prime to each other*. 9. Farther, let $7200m^2 = 4fg$, or let its factors be $2f$ and $2g$, supposing $s+r = 2f$, and $s-r = 2g$, we shall have $s=f+g$, and $r=f-g$; $f$ and $g$, also, must be prime to each other, and the one must be odd and the other even. Now, as $fg = 1800m^2$, we may resolve $1800m^2$ into two factors, the one being even and the other odd, and having at the same time no common divisor. 10. It is to be farther remarked, that since $r^2 = p^2 + q^2$, and since $r$ is a divisor of $p^2 + q^2$, $r=f-g$ must likewise be the sum of two squares (Art. 170); and as this number is odd, it must be contained in the formula $4n+1$. 11. If we now begin with supposing $m=1$, we shall have $fg = 1800 = 8 \times 9 \times 25$, and hence the following results: $f=1800$, and $g=1$, or $f=200$, and $g=9$, or $f=72$, and $g=25$, or $f=225$, and $g=8$. The 1st gives $r=f-g=1799$ (of form $4n+3$); 2d gives $r=f-g=191$ (of form $4n+3$); 3d gives $r=f-g=47$ (of form $4n+3$); 4th gives $r=f-g=217$ (of form $4n+1$); So that the first three must be excluded, and there remains only the fourth: from which we may conclude, generally, that the greater factor must be odd, and the less even; but even the value, $r=217$, cannot be admitted here, because that number is divisible by 7, which is not the sum of two squares†. 12. If $m=2$, we shall have $fg = 7200 = 32 \times 225$; for which reason we shall make $f=225$, and $g=32$, so that $r=f-g=193$; and this number being the sum of two squares, it will be worth while to try it. Now, as $q=120$, and $r=193$, and $p^2 = r^2 - q^2 = (r+q) \times (r-q)$, we shall have $r+q=313$, and $r-q=73$; but since these factors are not squares, it is evident that $p^2$ does not become a square. In the same manner, it would be in vain to substitute any other numbers for $m$, as we shall now shew. 230. *Theorem*. It is impossible for the two formulæ $p^2+q^2$, and $p^2+3q^2$, to be both squares at the same time; so that in the cases where one of them is a square, it is certain that the other is not. * Because $p$ is odd and $q$ is even; therefore $p^2+q^2=r^2$, and $p^2+3q^2=s^2$, must be both odd. B. † Because the sum of two squares, prime to each other, can only be divided by numbers of the same form. B.