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October 2023 - Solution
October 2023 Solution:
When engineering the problem we did not consider negative numbers and forgot to explicitly forbid them, making solutions such as -12866669 possible. We accepted those solutions since they are nice tricks (which doesn't help with the bonus part).
Here is a brief description of one method of solving the problem, given by Karl Mahlburg:
Regular version:
n = 100000000000000100009999999999999990000000000000010001000000000000001
* version:
n = 10^{2*s3} + 1 + 10^{s3+s2} + 10^{s3-s2} + 10^{s3+s1} + 10^{s3-s1} - 10^s3,
where [s1, s2, s3] = [415, 3819, 4234]
(A 8469-digit number!)
The regular version is a similar construction as the * version, just with S = [15, 19, 34]. The next smallest case that satisfies the problem is [24, 43, 67].
I found this construction in some of the links from the OEIS page for palindromic squares
Most relevantly, the Asymmetric (A) family from here
This shows that any S = [s1, s2, s1 + s2] will give an asymmetric integer w/ palindromic square, and a brief search starts returning examples that satisfy the congruence mod 1201.
The congruence mod 3565 = 5*23*31 is not much harder, since mod 5 is satisfied automatically, and 23 and 31 are small primes, so Chinese Remainder Theorem works out pretty easily (note my * version is a lift of the regular solution, since 10 has order 200 mod 1201).