Sol060625

Show that numbers of the form 4444...44 (2 $n$ 4's in a row) - 88...8 ( $n$ 8's in a row) are alway perfect squares.

$\mbox{Notice that }4444\dots 44 \,(2n\mbox{ times}) = \frac{4}{9}\left(10^{2n}-1\right)$

$\mbox{Notice that }8\dots 8 \,(n\mbox{ times}) = \frac{8}{9}\left(10^{n}-1\right)$

$\frac{4}{9}\left(10^{2n}-1\right)-\frac{8}{9}\left(10^{n}-1\right)= \frac{4}{9}\left(10^{2n} - 1 - 2\left(10^n-1\right)\right)=\left(\frac{2}{3}\left(10^n-1\right)\right)^2$

Retrieved from "http://www.efnet-math.org/w/Sol060625"