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 n2 digits. Since n2⋅(32+42)=n2⋅52 is a square number, we can take the number 33…344…4, where 3 and 4 appeared n2 times, to see that exists a number with that property for every 2n2-digit number. and for others cases I made some progress, but I could not figue out the case n=2019=3⋅673. I can't see why it would't exist.
Answer
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≤2019
Since 2019+3⋅2=2025=452, 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.
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