A repeating number is a number such that the first half of the digits are the same as the second half of the digits. For example, 2020 is a repeating number. What is the smallest repeating number that is also a square?

I found this 22 digit number: 
8264462810082644628100  = 90909090910^2

For n = 7, the run was 14 seconds.  Going to be too slow for larger numbers.

So I tried analytical.
For a repeat number of 2n digits, it must be divisible by a number that begins and ends with 1 and has (n-1) zeros in between (call it a 10..01 number).

If the factors of some 10..01 number are all unique, say a,b,c;  then the only way to make it into a square is a*b*c*a*b*c.  But then that square is a 10.2.01 number that has 2n+1 digits; that's one too many digits.

Find a 10..01 number containing two of the same prime factors, ie divisible by a square. So if the factors of the 10..01 number are, say, a,a,b,c then that 10..01 number times b times c will be a perfect square and will be divisible by the 10..01 number.  It may or may not have the correct number of digits, however.  If it has too few digits, then it can be multiplied by the smallest square of however many digits it takes to convert it to having 2n digits.

Note that all 10..01 numbers are 1 mod 4, and 2 mod 9, and 1 mod 25, so no square smaller than 6^2 will work.
I checked 10..01 numbers up to 10^13 + 1 and squares up to 2.5 million and found no examples of modulo 0 except for mod(10^11 + 1, 121)=0.

It turns out that 10^11 + 1 has the following list of prime factors:
    (11 * 11 * 23 * 4093 * 8779) = a*a*b*c*d
Now multiply that by b*c*d and (then possibly by another square to get 22 digits)  
82644628100826446281 = 11 * 11 * 23 * 4093 * 8779 * 23 * 4093 * 8779
which works if you were to add 2 leading zeros (it has only 20 digits)

So multiply by 100:  8264462810082644628100  has 22 digits.
11 * 11 * 23 * 4093 * 8779 * 23 * 4093 * 8779 * 100
This works and does not have leading zeros. 

Posted by Larry on 2020-09-09 09:36:19