Is there a number with $2019$ digits, all digits are positive, such that the sum of squares of its digits is a square number ?
The case is clear if the number has $n^2$ digits. Since $n^2\cdot(3^2+4^2)=n^2\cdot5^2$ is a square number, we can take the number $33\dots344\dots4$, where $3$ and $4$ appeared $n^2$ times, to see that exists a number with that property for every $2n^2$-digit number. and for others cases I made some progress, but I could not figue out the case $n=2019=3\cdot673$. I can't see why it would't exist.