Can $\,9\!\cdot\!10^n+4\,$ be a perfect square? I think $\,9\!\cdot\!10^n+4\,$ can be a perfect square, since it is $0 \pmod 4$ (a quadratic residue modulo $4$), and $1 \pmod 3$ (also a quadratic residue modulo  $3$).
But when I tried to find if $\;9\!\cdot\!10^n+4\,$ is a perfect square, I didn’t succeed. Can someone help me see if $\;9\!\cdot\!10^n+4\,$ can be a perfect square ?

Comment:
Lets try construction such a number. Suppose we have:
$k^2-9\times 10^n=4$
We use following known Pell's equation:
$x^2-Dy^1=1$
For $D=10$ we have $x=19$ and $y=6$ such that:
$19^2-10\times 6^2=1$
multiplying both sides by $2^2$ we get"
$38^2-10\times 12^2=4$
we rewrite this as:
$38^2-9\times(4^2\times 10)=4$
multiplying both sides by $25^2$ we get:
$(25\times 38)^2-9\times 10^5=4\times 25^2=2500=2496+4$
Or:
$[(25\times 38)^2-2496]-9\times 10^5=4$
Or generally:
$A=[(25\times 38)\times 10^m-(25\times 10^{2m}+96)]$
here in equation $k^2-9\times 10^n=4$, $n=2m+1$
$A$ must be perfect square.May be by brute force we can find such a number.
