# Is there a perfect square in the sequence $6, 96, 996, 9996, … , 99999…996, …$?

$6, 96, 996, 9996, ... , 99999...996, ...$

Consider the above number sequence. Here in the $n$ $^{th}$ term $n-1$ digits are $9$. How can we tell about the existence of a perfect square number in this sequence?

• If the perfect square were $10^{2n}-4$, then its factors would be $10^n+2,10^n-2$. So the perfect square must be of the form $10^{2n-1}-4$. – abiessu Mar 4 '16 at 17:57

The sum of the digits of the $n^{th}$ term of this sequence is $$\text{Sum}=9(n-1)+6=9n-3=3(3n-1)$$ Thus, every number in the sequence is divisible by $3$ but not by $9$. Hence, no number in the sequence can be a perfect square.
HINT: consider divisibility by a suitable small number $n$ and also by $n^2$