I know that $137 \times 73 = 10001 $. I am looking for a properly reasoned approach. I believe that there is some general result also which says that all numbers of the form $10^k + 1$ (for $k>2$) are NOT prime. Why? I mean can the general expression for the factors of such numbers be found?
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1if k is odd then the number is divisible by 11 – Mirko Nov 08 '16 at 19:32
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1If $k$ is 2 mod 4, then the number is divisible by 101. – Patrick Stevens Nov 08 '16 at 19:33
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I think there's a sieve thing going on here, with the binary expansion of $k$. – Patrick Stevens Nov 08 '16 at 19:34
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2In particular, if k is odd, then k = 11 * A095372(n). – Peter Kagey Nov 08 '16 at 19:34
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2@peter, I believe you wanted to say that $10^k + 1 = 11 \times A095372((k+1)/2)$ for an odd $k$. – Shraddheya Shendre Nov 08 '16 at 19:41
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2If $k$ has any odd factor $m>1$, $10^k+1$ is divisible by $10^{k/m}+1$. So that leaves only powers of $2$. – Robert Israel Nov 08 '16 at 19:44
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questions discussed elsewhere https://www.physicsforums.com/threads/primes-of-form-10-k-1.392807/ also http://mathforum.org/kb/message.jspa?messageID=7419673 – Mirko Nov 08 '16 at 19:52
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Since (as I mentioned in a comment) $k$ must be a power of $2$, you are asking whether the generalized Fermat number $F_n(10) = 10^{2^n}+1$ is prime for any $n > 2$. It is conjectured that for any $b$ there are only finitely many primes $F_n(b) = b^{2^n}+1$, and it is quite likely that there are none for $b=10$ with $n > 1$.
According to Wilfrid Keller's Prime factors of generalized Fermat numbers Fm(10) and complete factoring status , which I think has the latest status of the problem, it is still not known whether $F_{24}(10) = 10^{2^{24}}+1$ is prime.
Robert Israel
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