Lehmer’s Totient Conjecture

OPENMajorConjectureProposed 1932 · Full conjecture

Canonical statement

For n1n\ge1, let φ(n)=#(Z/nZ)×\varphi(n)=\#(\mathbb Z/n\mathbb Z)^\times be Euler’s totient. If n>1n>1 and φ(n)\varphi(n) divides n1n-1, then nn is prime.
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For \(n\ge1\), let \(\varphi(n)=\#(\mathbb Z/n\mathbb Z)^\times\) be Euler’s totient. If \(n>1\) and \(\varphi(n)\) divides \(n-1\), then \(n\) is prime.

Euler's totient satisfies φ(p)=p1\varphi(p)=p-1 for every prime pp, so the divisibility φ(n)n1\varphi(n)\mid n-1 holds trivially at primes. In 1932 D. H. Lehmer asked whether this property characterises the primes: if n>1n>1 and φ(n)\varphi(n) divides n1n-1, must nn be prime [Lehmer1932Totient]? The conjecture asserts that no composite number passes the test.

A composite counterexample is known to be tightly constrained: it would have to be odd and squarefree, and — since φ(n)n1\varphi(n)\mid n-1 forces an11(modn)a^{n-1}\equiv1\pmod n for every aa coprime to nn — it would in particular be a Carmichael number. Cohen and Hagis showed that such an nn must have many distinct prime factors [CohenHagis1980], and computational work such as Renze's has excluded counterexamples in very large ranges [Renze1998Lehmer].

The necessary conditions keep strengthening, but they neither exhaust the search space nor yield a contradiction. The conjecture remains open; settling it requires either a composite nn with φ(n)n1\varphi(n)\mid n-1 or a proof that the divisibility forces primality.

The boxed statement is the canonical open formulation — not a stronger variant or a related research program. The status reflects the catalog's last review; do your own literature search before investing serious effort.