Elements of Algebra

344 ELEMENTS PART II. $m^2 = \frac{a^2p^4+2ap^2q^2+q^4}{a^2p^4-2ap^2q^2+q^4}$ and $n^2 = \frac{4p^2q^2}{a^2p^4-2ap^2q^2+q^4}$; hence, $m^2-an^2 = \frac{a^2p^4+2ap^2q^2+q^4-4ap^2q^2}{a^2p^4-2ap^2q^2+q^4} = \frac{a^2p^4-2ap^2q^2+q^4}{a^2p^4-2ap^2q^2+q^4} = 1$. 83. We see, therefore, that the two numbers $m$ and $n$ must be such, that $m^2 = an^2 + 1$. So that, as $a$ is a known number, we must begin by considering the means of determining such an integer number for $n$, as will make $an^2 + 1$ a square; for then $m$ will be the root of that square; and when we have likewise determined the number $f$ so, that $af^2 + b$ may become a square, namely $g^2$, we shall obtain for $x$ and $y$ the following values in integer numbers; $x = ng - mf$, $y = mg - naf$; and thence, lastly, $ax^2 + b = y^2$. 84. It is evident, that having once determined $m$ and $n$, we may write instead of them $-m$ and $-n$, because the square $n^2$ still remains the same. But we have already shewn that, in order to find $x$ and $y$ in integer numbers, so that $ax^2 + b = y^2$, we must first know a case, such that $af^2 + b$ may be equal to $g^2$; when we have therefore found such a case, we must also endeavour to know, beside the number $a$, the values of $m$ and $n$, which will give $an^2 + 1 = m^2$: the method for which shall be described in the sequel, and when this is done, we shall have a new case, namely, $x = ng + mf$, and $y = mg + naf$, also $ax^2 + b = y^2$. Putting this new case instead of the preceding one, which was considered as known; that is to say, writing $ng + mf$ for $f$, and $mg + naf$ for $g$, we shall have new values of $x$ and $y$, from which, if they be again substituted for $x$ and $y$, we may find as many other new values as we please: so that, by means of a single case known at first, we may afterwards determine an infinite number of others. 85. The manner in which we have arrived at this solution has been very embarrassed, and seemed at first to lead us from our object, since it brought us to complicated fractions, which an accidental circumstance only enabled us to reduce: it will be proper, therefore, to explain a shorter method, which leads to the same solution. 86. Since we must have $ax^2 + b = y^2$, and have already found $af^2 + b = g^2$, the first equation gives us $b = y^2 - ax^2$, and the second gives $b = g^2 - af^2$; consequently, also, $y^2 - ax^2 = g^2 - af^2$, and the whole is reduced to determining the unknown quantities $x$ and $y$, by means of the known quantities $f$ and $g$. It is evident, that for this pur-