This question asks about concatenated squares to make a square number. For example $[4][9]=49, [16][9]=169, [3136][441]=3136441, [64][009]=64009$
I've been doing a bit of investigating for the case $[a][b]=10^na+b$ where $n$ is the number of digits of $b$ for fixed $n$. There seem to be plenty of solutions, though I treat cases where $b=0$ representing a string of zeros as redundant.
The case $n=2$ involves $x^2=100y^2+z^2$ where $1\leq z\leq 9$
I am looking for a proof or counterexample to the statement that $x=41, y=4, z=9$ leading to $1681=[16][81]$ is the only solution of this kind with $n=2$ - since there are no other solutions in my (finite) search space.
Note I've excluded cases like $8100=[81][00]$ - there are clearly infinitely many trivial solutions with two trailing zeros.
Could a proof be generalised to other $n$?
Its true that $\exists$ only one $x=41,y=4,z=9$.
$$x^2 = 100y^2 + z^2 \implies x^2 = (10y)^2 +z^2$$
Now, the problem reduces to finding all the pythogorean triplets whose smallest side is $\lt 10$ and the second smallest side is a multiple of $10$.
It can be seen that there are only six such triplets whose smallest side is $<10$
$(3,4,5),(5,12,13),(6,8,10),(7,24,25), (8,15,17), (9,40,41)$
of which only the last triplet has the second side multiple of $10$. Hence the only solution is $x=41,y=4,z=9$
This method can be used for all even $n$. For odd $n$ modify $10^ny^2 = 9.10^{n-1}y^2 + 10^{n-1}y^2$. In this case you need to find all pythogorean quadruples with corresponding conditions.