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Let $\sigma(k)$ denote the sum of all positive divisors of $k$.

Consider the equation $\sigma(n+1)=\sigma(n)+1$.

Has it been investigated before? (I did some search in books avaliable for me, and in the internet, didn't found)

Does it have a solution except $n=2$ ?

I checked up to $10^7$, there are no other solutions in this interval.

user957042
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    For this to be true, one of $\sigma(n),\sigma(n+1)$ must be odd. This question shows that this is so only if $n$ is a square or twice a square. That sounds like a good starting point. (At least, it makes computer search mush more efficient.) – saulspatz Aug 09 '21 at 20:33
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    The closest thing I know of is a result of Heath-Brown, which states that the equation $d(n)=d(n+1)$ has infinitely many solutions, where $d$ counts the number of divisors. Presumably, one might be able to adapt that method to the equation $\sigma(n) = \sigma(n+1)$. Unfortunately, I don't think the method would apply to this problem. – Joshua Stucky Aug 09 '21 at 21:49
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    As a side note, if the twin prime conjecture is true, then the equation $\sigma(n+2)=\sigma(n)+2$ has infinitely many solutions, since any prime $p$ with $p+2$ also prime would be a solution. – Joshua Stucky Aug 09 '21 at 21:52
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    Applying saulspatz's hint, I currently run a search for $n\le 10^{14}$. The only pairs I found so far with difference $1$ are $(2/3)$ (which is the solution mentioned in the question) and $(4/5)$ which is no solution because of $\sigma(4)>\sigma(5)$. – Peter Aug 10 '21 at 07:51
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    Finished. No more pairs found. This leads to a stronger conjecture. – Peter Aug 10 '21 at 08:02
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    @Peter What stronger conjecture does your last comment lead to? Or did you just mean your search "strengthens" in the sense of checking for much larger numbers? – coffeemath Aug 10 '21 at 08:24
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    @coffeemath The stronger conjecture is that the only pairs $(n/n+1)$ with $$|\sigma(n)-\sigma(n+1)|=1$$ are $(2/3)$ and $(4/5)$ – Peter Aug 10 '21 at 08:30

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