# Prove that all numbers $10^n + 1$ are square free

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.

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

• If one checks a lot more of those numbers, one finds 11, 21, 33, 39, 55, 63, .... – Jeppe Stig Nielsen Mar 13 '15 at 21:38
• @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. – kingW3 Mar 13 '15 at 21:51
• @kingW3 Thanks! Interesting. Forgive me for being naïve here, but how does one show that rigorously? – Mark Viola Mar 13 '15 at 22:01
• @Dr.MV $10^n+1\equiv2\mod3$. – Julián Aguirre Mar 13 '15 at 22:24
• $1^2\equiv1\mod3$, $2^2\equiv1\mod3$, $3^2\equiv0\mod3$. It wasn't so difficult, was it? – Julián Aguirre Mar 14 '15 at 7:33

$10^n+1$ 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 $5^n\gt\gt2^{n-1}$.

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