PROVED
This has been solved in the affirmative.
Are there infinitely many $n$ such that $\tau(n)=\tau(n+1)$, where $\tau$ is the divisor function?
A problem of Erdős and Mirsky
[ErMi52]. More generally, they ask about the estimation of the longest run of consecutive integers $\leq x$ which have the same number of divisors.
Spiro
[Sp81] proved that there are infinitely many $n$ such that $\tau(n)=\tau(n+5040)$, and Heath-Brown
[He84] improved her method to prove this for $\tau(n)=\tau(n+1)$. More generally, he showed that the number of such $n\leq x$ is\[\gg \frac{x}{(\log x)^7}.\]This lower bound was improved to\[\gg \frac{x}{(\log\log x)^3}\]by Hildebrand
[Hi87]. Erdős, Pomerance, and Sárközy
[EPS87] proved an upper bound of\[\ll\frac{x}{\sqrt{\log\log x}}.\]The true count is likely asymptotic to\[c\frac{x}{\sqrt{\log\log x}}\]for some constant $c>0$. This has been established for almost all $x$ by Tao and Teräväinen
[TaTe25].
See
[1003] for the analogous question with the Euler totient function.
This is discussed in problem B18 of Guy's collection
[Gu04].
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This page was last edited 02 December 2025.
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