Showing that a number is a perfect square

Question

How do I show that A=999...982000...081 is a square number when the oly information i have is that there are just as many 9s as there are 0s (with the number being greater than or equal to 4) ?

Answer 1

Actually the mathematical representation of your number is $$a_k=(10^k-1)\times10^{k+4}+82\times10^{k+2}+81$$ To show that this is a perfect square, let's get some hints:

If $a_k$ is to be equal to some $x^2$, then $\big((10^{k+2}-9)-9\big)\big((10^{k+2}-9)+9\big)=10^{k+2}(10^{k+2}-18)=(10^k-1)\times10^{k+4}+82\times10^{k+2}=x^2-81=(x-9)(x+9)$

So if this number is to be a perfect square, it is the perfect square of $10^{k+2}-9$.
Now use induction (base of the induction is $a_0=8281=(10^2-9)^2=91^2$) and go through the exact same procedure to prove that $a_k=(10^{k+2}-9)^2$