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Is this possible to prove? I've seen some patterns in the prime factorizations of these numbers, and would like to know if it is possible to prove this.

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No: $10^{11} + 1$ is divisible by $11^2$.

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    $\begingroup$ If one checks a lot more of those numbers, one finds 11, 21, 33, 39, 55, 63, .... $\endgroup$ – Jeppe Stig Nielsen Mar 13 '15 at 21:38
  • $\begingroup$ @Dr.MV It's the smallest number such that $10^n+1$ is divisible by perfect square in this instance it's divisible by $11^2$ when $n=11$ also on a side note $10^n+1$ can never be a perfect square. $\endgroup$ – kingW3 Mar 13 '15 at 21:51
  • $\begingroup$ @kingW3 Thanks! Interesting. Forgive me for being naïve here, but how does one show that rigorously? $\endgroup$ – Mark Viola Mar 13 '15 at 22:01
  • $\begingroup$ @Dr.MV $10^n+1\equiv2\mod3$. $\endgroup$ – Julián Aguirre Mar 13 '15 at 22:24
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    $\begingroup$ $1^2\equiv1\mod3$, $2^2\equiv1\mod3$, $3^2\equiv0\mod3$. It wasn't so difficult, was it? $\endgroup$ – Julián Aguirre Mar 14 '15 at 7:33
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$10^n+1$ is never a perfect square. $n$ must be odd. We have: $$10^n =c^2-1=(c+1)(c-1)$$ One of $(c+1)$ and $(c-1)$ must be a multiple of $10$, as otherwise we cannot introduce factors of $2$. And it must not have $100$ as a factor, otherwise the other number must be $2$.

and so w.l.o.g. we hsve say $c+1=2\times 5^n$ and $c-1=2^{n-1}$. But this is clearly impossible as $5^n\gt\gt2^{n-1}$.

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    $\begingroup$ A number can be square-free without being a perfect square. $\endgroup$ – anomaly Mar 18 '15 at 18:19
  • $\begingroup$ I think OP meant no squares not squarefree. $\endgroup$ – JMP Mar 18 '15 at 18:22

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