Elements of Algebra

CHAP. V. ADDITIONS. 539 Resolution of the Equation $AP^2 + BQ^2 = z^2$ in integer Numbers. 52. Supposing A greater than B, the equation will be written thus, $AP^2 = z^2 - BQ^2$, and as the numbers $p$, $q$, and $z$ must be integer, $z^2 - BQ^2$ must be divisible by A. Now, since A and $q$ are prime to each other (Art. 51), we shall, according to the method of Art. 48, make $z = nq - Aq'$, $n$ and $q'$ being two indeterminate integers; which will change the formula, $z^2 - BQ^2$, into $(n^2 - B)q^2 - 2n Aqq' + A^2q'^2$, in which $n^2 - B$ must be divisible by A, taking for $n$ an integer number, not $> \frac{A}{2}$. We shall try therefore for $n$ all the integer numbers that do not exceed $\frac{A}{2}$, and if we find none that makes $n^2 - B$ divisible by A, we conclude immediately, that the equation $AP^2 = z^2 - BQ^2$ is not resolvible in whole numbers, and therefore that the quantity $Ay^2 + B$ can never become a square. But if we find one or more satisfactory values of $n$, we must substitute them, one after the other, for $n$, and proceed in the calculation, as shall now be shewn. I shall only remark farther, that it would be useless to give $n$ values greater than $\frac{A}{2}$, for calling $n'$, $n''$, $n'''$, &c. the values of $n$ less than $\frac{A}{2}$, which will render $n^2 - B$ divisible by A, all the other values of $n$ that will have the same effect will be contained in these formulæ, $n' \pm \mu' A$, $n'' \pm \mu'' A$, $n''' \pm \mu''' A$, &c. (Chap. IV. 47). Now, substituting these values for $n$, in the formula, $(n^2 - B)q^2 - 2n Aqq' + A^2q'^2$, that is to say, $(nq - Aq')^2 - Bq^2$, it is evident that we shall have the same results, as if we only put $n'$, $n''$, $n'''$, &c. instead of $n$, and added to $q'$ the quantities $\mp \mu' q$, $\mp \mu'' q$, $\mp \mu''' q$, &c. so that, as $q'$ is an indeterminate number, these substitutions would not give formulæ different from what we should have, by the simple substitution of the values $n'$, $n''$, $n'''$, &c. 53. Since, therefore, $n^2 - B$ must be divisible by A, let A'