By observations, I found that the numbers 11, 101, 1001, 10001, 100001, 1000001, 10000001 and up to 11 digits are all made up of completely unique prime factorisations (square free). E.g. 1001 is equal to 7*11*13, 3 of them are unrepeated prime numbers. Is there a method, possibly using discrete mathematics and concepts of numbers to prove or at least generalize the assumption that 10000...1 is also composed of unique prime factors (a1,a2,...,an) such that they do not equate one another, Or is this assumption unable to be made?
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1$\begingroup$ FYI, a number without repeated prime factors is known as square-free, in that it's not divisible by any square number greater than $1$. $\endgroup$– Theo BenditMar 25, 2018 at 1:13
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$\begingroup$ 'xxxxxxxxxxxxxxxxxxxxxxxx $\endgroup$– William ElliotMar 25, 2018 at 1:53
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$$10^{11}+1 = 11^2 \cdot 23 \cdot 4093 \cdot 8779$$
answered Mar 25, 2018 at 1:17
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1$\begingroup$ So $10^{11} \equiv -1 \pmod{11^2}$, and therefore $11^2 \mid 10^{11(2n+1)}+1\,$ for all $\,n \in \mathbb{N}\,$. $\endgroup$– dxivMar 25, 2018 at 1:23