Can anyone help me prove this question? Thank you.

Prove that there exists a positive integer $n$ such that the number $2^n+3^n$ has exactly $23$ prime divisors.

(I have thought about some fomulas about $\omega (n)$, like ${\displaystyle \sum _{d\mid n}|\mu (d)|=2^{\omega (n)}}$, or construct the sequence ${n_k}$ such that $2^{n_{k+1}}+3^{n_{k+1}}$ has exactly one more prime divisor than $2^{n_{k}}+3^{n_{k}}$, but it seems to be hard.)

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    $\begingroup$ Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or closed. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. $\endgroup$ Apr 11, 2020 at 7:46
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    $\begingroup$ Does n have to be integer or whole number or can it just be rational? In which case, this is trivial. $\endgroup$ Apr 11, 2020 at 8:00
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    $\begingroup$ Good question. I cannot solve it, but I found that $2^{3^k}+3^{3^k}$ has more than $k$ prime divisors. Quick proof: Use induction. $2^{3^{k+1}}+3^{3^{k+1}}=(2^{3^k}+3^{3^k})(2^{3^{2k}}-2^{3^k}3^{3^k}+3^{3^{2k}})$ and easy calculation shows that $2^{3^k}+3^{3^k}$ and $2^{3^{2k}}-2^{3^k}3^{3^k}+3^{3^{2k}}$ are coprime. $\endgroup$
    – flowsignal
    Apr 11, 2020 at 16:33
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    $\begingroup$ I wish "23" were "13", as then I could just say "Observe $2^{150}+3^{150}=13\cdot61\cdot2341\cdot4621\cdot4801\cdot11701\cdot24001\cdot797701\cdot9802501\cdot104189401\cdot1333073701\cdot2296284901\cdot12826646101$.". $\endgroup$
    – Mark S.
    Apr 12, 2020 at 16:34
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    $\begingroup$ @MarkS. I think 23 is a "special" number, but I don't know why $\endgroup$
    – Mergolyx
    Apr 13, 2020 at 1:34

1 Answer 1


Proof by example:


Factors into exactly $23$ distinct prime factors:

$$ 13, 37, 61, 73, 181, 229, 2053, 8209, 13681, 15277, 25309, 102829, 30883969, 196498153, 724174057, 743780461, 2117021041, 2230888573, 54458107801, 8077765456081, 80381675102807053, 103911691734684541, 324469548901114381 $$

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    $\begingroup$ How did you find this factorization? $\endgroup$
    – lhf
    Apr 16, 2020 at 11:45
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    $\begingroup$ @lhf I was spending $10$ seconds at most factoring a single $2^n+3^n$ expression (discarding the case if a single factor wasn't found in that time frame), for all $n$ until an example was found, in Wolfram Mathematica. $\endgroup$
    – Vepir
    Apr 16, 2020 at 11:50

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