Is there a 2019-digit number such that the sum of squares of its digits is a square number? 
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.
 A: Let $a$ count the number of $1$'s in your $2019$-digit number, $b$ the number of $2$'s, $c$ the number of $3$'s, etc., up to $h$ the number of $8$'s and $k$ the number of $9$'s.  Then the sum of the squares of the digits is
$$a+4b+9c+16d+25e+36f+49g+64h+81k$$
with
$$a+b+c+d+e+f+g+h+k=2019$$
So all we need is for
$$2019+3b+8c+15d+24e+35f+48g+63h+80k$$
to be a square with $b+c+d+e+f+g+h+k\le2019$
Since $2019+3\cdot2=2025=45^2$, and since non-negative combinations of $3b+8c$ form every integer greater than $13$ all by themselves, you have to go pretty high up to find a square that is not expressible as the sum of the squares of $2019$ nonzero digits. It might be of interest to find the first square that gets missed.
A: Note that $45^2 = 2025$ and $2017(1^2) + 2(2^2) = 2025$.
In general, this will work for a lot of numbers, because, by changing say $n$ $1$'s to $2$'s and $m$ $1$'s to $3$'s, we can add $3n + 8m$ to our number. Setting $m$ and $n$ to whatever gives us a lot of possibilities
A: $2015\cdot1^2+5^2+7^2+2^2+4^2=2116=46^2$
A: Since I saw answers with sum of digits squared $45^2$ and $46^2$,
I found  $47^2=2015\cdot1^2+4^2+4^2+9^2+9^2$
A: Try $(10^{2019}-1)/9+11$.  Can you identify which number squared is the digital sum for this case?
