We have:
$$1=1^2$$ and $$25=5^2=3^2+4^2$$
and $$441=21^2=20^2+4^2+5^2$$
So, for $k=1,2,3$ we have a $k$-digit number that is a perfect square and a sum of $k$ different non-zero perfect squares.
I did not seek further but this observation led me to conjecture:
For every $k \in \mathbb N$ there exists a $k$-digit number that is a perfect square and a sum of exactly $k$ different non-zero perfect squares. Is this known to be true/false?
Thanks for the help.
Small values of $k$ are simple to check and every natural number is the sum of four squares by Lagrange's theorem. Asymptotics for the average order of the representation function $$ r_k(n)=\left|\{(a_1,\ldots,a_k)\in\mathbb{Z}^k : a_1^2+\ldots+a_k^2 = n\}\right| $$ are well-known thanks to Hardy's earliest works (we are essentially counting the lattice points inside a hypersphere, which is a direct generalization of Gauss circle problem) and the constraint about different squares has a little impact. So yes, it is just a matter of recalling some well-known facts.
One may also use the Rusza-Plunnecke inequality since the set of squares is an additive basis of order $4$; or show that there are infinite rational points on the hypersphere $x_1^2+\ldots+x_k^2 = 1$, which is straightforward through the usual methods of algebraic geometry recalling Vieta jumping.