Elements of Algebra

426 ELEMENTS PART II. namely, by making $p = 13$, and $q = 2$; for the first becomes $169 - 120 = 49$, and the second $169 + 120 = 289$. 3. Likewise the two formulæ $p^2 - 15q^2$, and $p^2 + 15q^2$; for if we make $p = 17$, and $q = 4$, we have, for the first, $289 - 240 = 49$, and for the second $289 + 240 = 529$. 4. The two formulæ $p^2 - 5q^2$, and $p^2 + 5q^2$, become likewise squares: namely, when $p = 41$, and $q = 12$; for then $p^2 - 5q^2 = 1681 - 720 = 961 = 31^2$, and $p^2 + 5q^2 = 1681 + 720 = 2401 = 49^2$. 5. The two formulæ $p^2 - 7q^2$, and $p^2 + 7q^2$, are squares, if $p = 337$, and $q = 120$; for the first is then $113569 - 100800 = 12769 = 113^2$, and the second is $113569 + 100800 = 214369 = 463^2$. 6. The formulæ $p^2 - 14q^2$, and $p^2 + 14q^2$, become squares in the case of $p = 65$, and $q = 12$; for then $p^2 - 14q^2 = 4225 - 2016 = 2209 = 47^2$, and $p^2 + 14q^2 = 4225 + 2016 = 6241 = 79^2$. 227. Example 2. When the two numbers $m$ and $n$ are in the ratio of $1$ to $2$; that is to say, when $a = 1$, and $b = 2$, and therefore $m = z$, and $n = 2z$, to find such values for $z$, that the formulæ $p^2 + zq^2$ and $p^2 + 2zq^2$ may be transformed into squares. Here it would be superfluous to make use of the general formulæ already given, since this example may be immediately reduced to the preceding. In fact, if $p^2 + zq^2 = r^2$, and $p^2 + 2zq^2 = s^2$, we have, from the first equation, $p^2 = r^2 - zq^2$; which being substituted in the second, gives $r^2 + zq^2 = s^2$; so that the question only requires, that the two formulæ, $r^2 - zq^2$, and $r^2 + zq^2$, may become squares; and this is evidently the case of the preceding example. We shall consequently have for $z$ the following values: $6, 30, 15, 5, 7, 14, \ldots$. We may also make a similar transformation in a general manner. For, supposing that the two formulæ $p^2 + mq^2$, and $p^2 + nq^2$, may become squares, let us make $p^2 + mq^2 = r^2$, and $p^2 + nq^2 = s^2$; the first equation gives $p^2 = r^2 - mq^2$; the second will become $s^2 = r^2 - mq^2 + nq^2$, or $r^2 + (n-m)q^2 = s^2$: if, therefore, the first formulæ are possible, these last $r^2 - mq^2$, and $r^2 + (n-m)q^2$, will be so likewise; and as $m$ and $n$ may be substituted for each other, the formulæ $r^2 - nq^2$, and $r^2 + (m-n)q^2$, will also be possible: on the contrary, if the first are impossible, the others will be so likewise. 228. Example 3. Let $m$ be to $n$ as $1$ to $3$, or let $a = 1$, and $b = 3$, so that $m = z$, and $n = 3z$, and let it be required to transform into squares the formulæ $p^2 + zq^2$, and $p^2 + 3zq^2$.