Elements of Algebra

358 ELEMENTS. PART II. wherefore $n=ep+\sqrt{(e^2p^2-p^2+1)}$. Now, the irrationality disappeared by supposing $p=1$; so that $n=2e$, and $m=2e^2-1$. This also is evident; for, since $a=e^2-1$, and $n=2e$, we find $an^2+1=4e^4-4e^2+1$, or equal to the square of $2e^2-1$. For example, let $a=24$, or $e=5$, we shall have $n=10$, and $24n^2+1=2401=(49)^2$. 110. Let us now suppose $a=e^2+1$, or a greater than a square by unity. Here we must have $(e^2+1)n^2+1=m^2$, and $m$ will evidently be greater than $en$. Let us, therefore, write $m=en+p$, and we shall have $(e^2+1)n^2+1=e^2n^2+2enp+p^2$, or $n^2=2enp+p^2-1$; whence $n=ep+\sqrt{(e^2p^2+p^2-1)}$. Now, we may make $p=1$, and shall then have $n=2e$; therefore $m=2e^2+1$; which is what ought to be the result from the consideration, that $a=e^2+1$, and $n=2e$, which gives $an^2+1=4e^4+4e^2+1$, the square of $2e^2+1$. For example, let $a=17$, so that $e=4$, and we shall have $17n^2+1=m^2$; by making $n=8$, and $m=33$. 111. Lastly, let $a=e^2+2$, or greater than a square by 2. Here, we have $(e^2+2)n^2+1=m^2$, and, as before, $m>en$; therefore we shall suppose $m=en+p$, and shall thus have $e^2n^2+2n^2+1=e^2n^2+2enp+p^2$, or $2n^2=2epn+p^2-1$, which gives $n=\frac{ep+\sqrt{(e^2p^2+2p^2-2)}}{2}$. Let $p=1$, we shall find $n=e$, and $m=e^2+1$; and, in fact, since $a=e^2+2$, and $n=e$, we have $an^2+1=e^4+2e^2+1$, which is the square of $e^2+1$. For example, let $a=11$, so that $e=3$; we shall find $11n^2+1=m^2$, by making $n=3$, and $m=10$. If we * In this case, likewise, the radical sign vanishes, if we make $p=0$: and this supposition incontestably gives the least possible numbers for $m$ and $n$, namely, $n=1$, and $m=e$; that is to say, if $e=5$, the formula $24n^2+1$ becomes a square by making $n=1$; and the root of this square will be $m=e=5$. F. T.