Elements of Algebra

CHAP. XIV. OF ALGEBRA. 427 Since $a = 1$, and $b = 3$, the question will be possible in all the cases in which $zq^2 = 4vy(v + y) \times (v + 3y)$, and $p = v^2 - 3y^2$. Let us therefore adopt the following values for $v$ and $y$: $v$ | 1 | 3 | 4 | 1 | 16 --|---|---|---|---|--- $y$ | 1 | 2 | 1 | 8 | 9 $v+y$ | 2 | 5 | 5 | 9 | 25 $v+3y$ | 4 | 9 | 7 | 25 | 43 $zq^2$ | $16 \times 2$ | $4 \times 9 \times 30$ | $4 \times 4 \times 35$ | $4 \times 9 \times 25 \times 4 \times 2$ | $4 \times 9 \times 16 \times 25 \times 43$ $q^2$ | 16 | $4 \times 9$ | $4 \times 4$ | $4 \times 9 \times 25$ | $4 \times 9 \times 16 \times 25$ $z$ | 2 | 30 | 35 | 2 | 43 $p$ | 2 | 3 | 13 | 191 | 13 Now, we have here two cases for $z = 2$, which enables us to transform, in two ways, the formulæ $p^2 + 2q^2$, and $p^2 + 6q^2$. The first is, to make $p = 2$, and $q = 4$, and consequently also $p = 1$, and $q = 2$; for we have then from the last $p^2 + 2q^2 = 9$, and $p^2 + 6q^2 = 25$. The second is, to suppose $p = 191$, and $q = 60$, by which means we shall have $p^2+2q^2 = (209)^2$, and $p^2+6q^2 = (241)^2$. It is difficult to determine whether we cannot also make $z = 1$; which would be the case, if $zq^2$ were a square: but, in order to determine the question, whether the two formulæ $p^2 + q^2$, and $p^2 + 3q^2$, can become squares, the following process is necessary. 229. It is required to investigate, whether we can transform into squares the formulæ $p^2 + q^2$, and $p^2 + 3q^2$, with the same values of $p$ and $q$. Let us here suppose $p^2+q^2=r^2$, and $p^2+3q^2=s^2$, which leads to the investigation of the following circumstances. 1. The numbers $p$ and $q$ may be considered as prime to each other; for if they had a common divisor, the two formulæ would still continue squares, after dividing $p$ and $q$ by that divisor. 2. It is impossible for $p$ to be an even number; for in that case $q$ would be odd; and, consequently, the second formula would be a number of the class $4n + 3$, which cannot become a square; wherefore $p$ is necessarily odd, and $p^2$ is a number of the class $8n + 1$. 3. Since $p$ therefore is odd, $q$ must in the first formula not only be even, but divisible by 4, in order that $q^2$ may become a number of the class $16n$, and that $p^2 + q^2$ may be of the class $8n + 1$. 4. Farther, $p$ cannot be divisible by 3; for in that case, $p^2$ would be divisible by 9, and $q^2$ not; so that $3q^2$ would