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I have already posted this question with username "mamihlapinatapai", but apparently I've made a major/big error to my question. I couldn't edit it because my question has been answered by two users.

My previous statement that $16^n+1$ can be a semiprime if and only if n is a prime is FALSE, the correct statement is: $16^n+1$ can be a semiprime if and only if n is either a prime, a power of 2, or the product of a power of two with an odd prime.

I also seriously misinterpreted Mr. Norata's conjecture ( I hope Mr. Norata will forgive me). Mr. Norata said $16^p+1$ and not $16^n+1$ ( I misinterpreted it as $16^n+1$). So, Mr. Norata conjecture that $16^{317}+1$ is the last/largest semiprime of the form $16^p+1$, where P is a prime. I've checked p up to 10009 and apparently the conjecture is still hold(!).

Can you prove or disprove Mr. Norata's conjecture ?

iadvd
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Mika Setiawan
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1 Answers1

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The conjecture is false, because 30197 is prime and $16^{30197} + 1$ is semiprime. See https://oeis.org/A057182

jbuddenh
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