Erdős Problem #1055
No proof or disproof is known, and it isn't known whether a finite computation could settle it.
What the community database records, not our verdict. erdosproblems.com no longer shows statuses (why).
A prime \(p\) is in class \(1\) if the only prime divisors of \(p+1\) are \(2\) or \(3\). In general, a prime \(p\) is in class \(r\) if every prime factor of \(p+1\) is in some class \(\leq r-1\), with equality for at least one prime factor. The classes partition the primes, so `IsOfClass r p` says that \(p\) is in class *exactly* \(r\): the first conjunct of the successor case rules out every class \(\leq r - 1\) for \(p\) itself. Without it `IsOfClass 2` would also hold of every prime of class \(1\), because at \(r = 2\)
A prime \(p\) is in class \(1\) if the only prime divisors of \(p+1\) are \(2\) or \(3\). In general, a prime \(p\) is in class \(r\) if every prime factor of \(p+1\) is in some class \(\leq r-1\), with equality for at least one prime factor. Let \(p_r\) is the least prime in class \(r\).
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modulo Dirichlet every class is nonempty and infinitely many primes lie at or above each class; an independent exact certificate gives \(p_1,\ldots,p_{18}\), while exact-class infinitude and the behav
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