Prove that a number composed of only $2n$ "1"s, minus another number composed of only $n$ "2"s, is a perfect square.
The number composed of $n$ ones is $(10^n-1)/9$. So your number is $$\frac{10^{2n}-1}9-2\frac{10^n-1}9=\frac{a^2-1-2(a-1)}9$$ with $a=10^n$. I'll leave you to carry on...