Let's start with some calculations: $$7^{2} = 49, 97^{2} = 9409, 997^{2} = 994009, \dots$$I proved that $A = B ^ {2}, B = 9\dots97$ is $supersquare$ Let's assume that $B$ has $x$ digits $9$ in the front. $\implies B = 10^{x+1}-3 \implies A = B^{2} = (10^{x+1}-3)^{2} = 10^{2x+2}-6\cdot10^{x+1}+9 = 9\dots940\dots09$ where we have exactly $x$ time $0$'s and $9$'a in $A$. The only requirement for $x$ is $x \in N \implies$ inflinity number of $supersquares$.