My question is somewhat related to this question:
Prove that all numbers $10^n + 1$ are square free
Just with a little modification. I want to find the values of $n$ for which a number of this form is not square-free. One value suggested in the above link is $n = 11$. Are there any other values of $n$ for which the number is not square-free.
Additional question: Are there infinitely many values of $n$ which are an answer to my question?
On
The values are in a list that runs as follows, and all odd multiples thereof.
11, 21, 39, 202, 243, 253, 292, ...
which are multiples of the squares of 11, 7, 13, 101, 487, 253, 73, ...
Here the rule is simple. 11 divides 10+1, 10^3+1, 10^5+1, &c, and thus 11^2 divides 11 times the index, viz 11^11+1, 11^33+1, 11^55+1, &c.
Any prime p that has a period 2n, divides 10^(2kn+n)+1. This represents that p² has a period also, and it is p times the period of p. 2k+1 represents an odd number.
The example of 487 dividing 10^243+1, is because 487 is a decimal sevenite, that is, a prime that divides its own period in base 10. The prime example of a sevenite is in base 18, where
1/7 = 0. 2 10 5 2 10 5 ... but 2 10 5 is a multiple of 7 1/7^2 = 0. 0 6 11 0 6 11 0 6 11 .. but 6 11 is a multiple of 7 1/7^3 = 0. 0 0 17 0 0 17 0 0 17 ... and 17 is not a multiple of 7
In the example above, every odd multiple of 2 in 18^n+1, is a multiple of 5, and every occasion where 18 divides n an odd number of times, it is also a multiple of 37.
Yes, there are infinitely many. In fact, it can be proved that if $10^n+1$ is not square free and $k$ is odd, then $10^{nk}+1$ is also not square fre. This follows from the equality$$10^{nk}+1=10^{nk}+1^k=(10^n)^k-(-1)^k$$and from the fact that $10^n+1$ is not square free.
You can learn more about these numbers at the OEIS.