Przekształćmy postać podanej liczby:

$\underset{10\text{cyfr}}{\underbrace{111\dots 1}}0\underset{10\text{cyfr}}{\underbrace{888\dots 8}}9=111\dots 1\cdot {10}^{12}+888\dots 8\cdot 10+9=111\dots 1\cdot {10}^{12}+8\cdot 111\dots 1\cdot 10+9$  

• Zauważmy, że

$111..1=1\cdot {10}^0+1\cdot {10}^1+1\cdot {10}^2+\dots +1\cdot {10}^9=1\cdot \left(1+10+{10}^2+\dots +{10}^9\right)=1+10+{10}^2+\dots +{10}^9=1\cdot \frac{1-{10}^{10}}{1-10}=\frac{{10}^{10}-1}{9}$  

stąd:

 

$={10}^{12}\cdot \frac{{10}^{10}-1}{9}+80\cdot \frac{{10}^{10}-1}{9}+9=\frac{1}{9}\cdot \left({10}^{12}\cdot \left({10}^{10}-1\right)+80\cdot \left({10}^{10}-1\right)+81\right)=\frac{{10}^{22}-{10}^{12}+80\cdot {10}^{10}-80+81}{9}=\frac{{10}^{22}-10\cdot {10}^{11}+8\cdot {10}^{11}+1}{9}=\frac{{10}^{22}-2\cdot {10}^{11}+1}{9}=\frac{{\left({10}^{11}\right)}^2-2\cdot {10}^{11}\cdot 1+1}{9}={\left[\frac{{10}^{11}-1}{3}\right]}^2$  

a więc:

$\underset{10\text{cyfr}}{\underbrace{111\dots 1}}0\underset{10\text{cyfr}}{\underbrace{888\dots 8}}9={\left[\frac{{10}^{11}-1}{3}\right]}^2$