The question is equivalent to finding integer m such that 10^m+1 = a^2*b, where a, b are integers with a > 1 and b < 10^m+1. Then the number we want is (10^m+1)*b = a^2*b^2 = (ab)^2.
I don’t think there is any general format for m. For example, 10^11+1 is divisible by 11^2, 10^21+1 is divisible by 7^2, and 10^39+1 is divisible by 13^2.
An interesting case is when m is odd integer (m = 2k+1). 10^m+1 is divisible by 11, and the quotient is 90…9091, where “90” appears k-1 times. With proper k, 90…9091 can be divisible by 11, i.e., 9k=1 (mod 11). So k = 5+11n, that is m = 11+22n, n = 0, 1, 2, …
When n = 0, the number we find is (10^11+1)/11^2=826446281. |
Good job. There is a minor correction. For (10^m + 1) * k to be a repeat number, k must have m digits. If 10^m + 1 = a^2 * b, a must be >= 7, and therefore the number we want is (10^m + 1) * b * c^2, where 0.1 < c^2 / a^2 < 1. This is what I call a general formula, although it is not general enough because it does not say how to find m. There is probably no general format for m, but it looks like for any odd a which is not a multiple of 3 or 5, m exists. Using the above general formula, the smallest repeat square I found is a repeat of 826446281*16, or 1322314049613223140496. |
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