Elements of Algebra

CHAP. XIV. OF ALGEBRA. 423 $(r^2 + 2s^2)^2$. The latter formula also becomes $p^2 + 6q^2 = r^4 + 20r^2s^2 + 4s^4$; and we know a case in which it becomes a square, namely, when $p=1$, and $q=2$, which gives $r=1$, and $s=1$; or, generally, $r=s$; so that the formula is $25s^4$. Knowing this case, therefore, let us make $r=s+t$; and we shall then have $r^2 = s^2 + 2st + t^2$, or $r^4 = s^4 + 4s^3t + 6s^2t^2 + 4st^3 + t^4$; so that our formula will become $25s^4 + 44s^3t + 26s^2t^2 + 4st^3 + t^4$: and, supposing its root to be $5s^2 + fst + t^2$, we shall make it equal to the square $25s^4 + 10fs^3t + f^2s^2t^2 + 10s^2t^2 + 2fst^3 + t^4$, by which means the first and last terms will be destroyed. Let us likewise make $2f = 4$, or $f=2$, in order to remove the last terms but one, and we shall obtain the equation $44s + 26t = 10fs + 10t + f^2t = 20s + 14t$, or $2s = -t$, and $s/t = -1/2$; therefore $s=-1$, and $t=2$, or $t=-2s$; and, consequently, $r=-s$, also $r^2=s^2$, which is nothing more than the case already known. Let us rather, therefore, determine $f$ in such a manner, that the second terms may vanish. We must make $10f=44$, or $f = 22/5$; and then dividing the other terms by $st^2$, we shall have $26s + 4t = 10s + f^2s + 2ft$, that is, $-84/5 s = 24/5 t$; which gives $t = -7/10 s$, and $r=s+t=3/10 s$, or $r/s = 3/10$; so that $r=3$, and $s=10$; by which means we find $p=2s^2-r^2=191$, and $q=2rs=60$, and our formulæ will be $p^2 + 2q^2 = 43681 = (209)^2$ and $p^2 + 6q^2 = 58081 = (241)^2$. 224. *Remark*. In the same manner, other numbers may be found for $m$ and $n$, that will make our formulæ squares; and it is proper to observe, that the ratio of $m$ to $n$ is arbitrary. Let this ratio be as $a$ to $b$, and let $m=az$, and $n=bz$; it will be required to know how $z$ is to be determined, in order that the two formulæ $p^2 + azq^2$, and $p^2 + bzq^2$, may be transformed into squares; the method of doing which we shall explain in the solution of the following problem. 225. *Question 9*. Two numbers, $a$ and $b$, being given, to find the number $z$ such, that the two formulæ, $p^2 + azq^2$, and $p^2 + bzq^2$, may become squares; and, at the same time, to determine the least possible values of $p$ and $q$. Here, if we make $p^2 + azq^2 = r^2$, and $p^2 + bzq^2 = s^2$, and multiply the first equation by $a$, and the second by $b$, the difference of the two products will furnish the equation