I am studying Giuga numbers and I am reading a paragraph that says all Giuga numbers $n$ satisfy the property $\left(\sum_{p|n}1/p\right) - 1/n \in \mathbb{N}$ and beside it says that all known Giuga numbers also satisfy $\sum_{p|n}{1}/{p} - 1/n = 1$. So is it proven or just a conjecture?
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3$\begingroup$ I've fixed your English (and formatting) which has many mistakes (this is not a problem — typically users here do such corrections), I'm not sure why you rolled back. $\endgroup$– YCorCommented Nov 3, 2023 at 16:52
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$\begingroup$ I reinstated your corrections, @YCor . $\endgroup$– Carlo BeenakkerCommented Nov 3, 2023 at 16:55
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$\begingroup$ @YCor sorry I didn't knew what rollback I thought it will apply the change but it didn't happen $\endgroup$– Raj Pratap SinghCommented Nov 3, 2023 at 17:29
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$\begingroup$ Note that $\sum_{p\mid n}(1/p)>2$ implies $n$ is divisible by at least 30 different primes, so the kind of $n$ you ask about, if it exists, is going to be rather large. $\endgroup$– Gerry MyersonCommented Nov 4, 2023 at 9:01
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1 Answer
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The prime factors of a Giuga number are all distinct; hence the equality in the OP is equivalent to $$\sum_{p|n}1/p - \prod_{p|n}1/p =\nu\in \mathbb{N},$$ which was proven by Giuga in his original paper (as stated in Borwein et al.). The 13 known Giuga numbers all satisfy $\nu=1$, it is unknown whether this is true for all Giuga numbers (see OEIS).
answered Nov 3, 2023 at 16:38