Agoh–Giuga conjecture

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In number theory the Agoh–Giuga conjecture on the Bernoulli numbers Bk postulates that p is a prime number if and only if

p B p − 1 ≡ − 1 ( mod p ) . {\displaystyle pB_{p-1}\equiv -1{\pmod {p}}.} {\displaystyle pB_{p-1}\equiv -1{\pmod {p}}.}

It is named after Takashi Agoh and Giuseppe Giuga.

Equivalent formulation

The conjecture as stated above is due to Takashi Agoh (1990); an equivalent formulation is due to Giuseppe Giuga, from 1950, to the effect that p is prime if and only if

1 p − 1 + 2 p − 1 + ⋯ + ( p − 1 ) p − 1 ≡ − 1 ( mod p ) {\displaystyle 1^{p-1}+2^{p-1}+\cdots +(p-1)^{p-1}\equiv -1{\pmod {p}}} {\displaystyle 1^{p-1}+2^{p-1}+\cdots +(p-1)^{p-1}\equiv -1{\pmod {p}}}

which may also be written as

∑ i = 1 p − 1 i p − 1 ≡ − 1 ( mod p ) . {\displaystyle \sum _{i=1}^{p-1}i^{p-1}\equiv -1{\pmod {p}}.} {\displaystyle \sum _{i=1}^{p-1}i^{p-1}\equiv -1{\pmod {p}}.}

It is trivial to show that p being prime is sufficient for the second equivalence to hold, since if p is prime, Fermat's little theorem states that

a p − 1 ≡ 1 ( mod p ) {\displaystyle a^{p-1}\equiv 1{\pmod {p}}} {\displaystyle a^{p-1}\equiv 1{\pmod {p}}}

for a = 1 , 2 , … , p − 1 {\displaystyle a=1,2,\dots ,p-1} {\displaystyle a=1,2,\dots ,p-1}, and the equivalence follows, since p − 1 ≡ − 1 ( mod p ) . {\displaystyle p-1\equiv -1{\pmod {p}}.} {\displaystyle p-1\equiv -1{\pmod {p}}.}

Status

The statement is still a conjecture since it has not yet been proven that if a number n is not prime (that is, n is composite), then the formula does not hold. It has been shown that a composite number n satisfies the formula if and only if it is both a Carmichael number and a Giuga number, and that if such a number exists, it has at least 13,800 digits (Borwein, Borwein, Borwein, Girgensohn 1996). Laerte Sorini, in a work of 2001 showed that a possible counterexample should be a number n greater than  1036067 which represents the limit suggested by Bedocchi for the demonstration technique specified by Giuga to his own conjecture.

Relation to Wilson's theorem

The Agoh–Giuga conjecture bears a similarity to Wilson's theorem, which has been proven to be true. Wilson's theorem states that a number p is prime if and only if

( p − 1 ) ! ≡ − 1 ( mod p ) , {\displaystyle (p-1)!\equiv -1{\pmod {p}},} {\displaystyle (p-1)!\equiv -1{\pmod {p}},}

which may also be written as

∏ i = 1 p − 1 i ≡ − 1 ( mod p ) . {\displaystyle \prod _{i=1}^{p-1}i\equiv -1{\pmod {p}}.} {\displaystyle \prod _{i=1}^{p-1}i\equiv -1{\pmod {p}}.}

For an odd prime p we have

∏ i = 1 p − 1 i p − 1 ≡ ( − 1 ) p − 1 ≡ 1 ( mod p ) , {\displaystyle \prod _{i=1}^{p-1}i^{p-1}\equiv (-1)^{p-1}\equiv 1{\pmod {p}},} {\displaystyle \prod _{i=1}^{p-1}i^{p-1}\equiv (-1)^{p-1}\equiv 1{\pmod {p}},}

and for p=2 we have

∏ i = 1 p − 1 i p − 1 ≡ ( − 1 ) p − 1 ≡ 1 ( mod p ) . {\displaystyle \prod _{i=1}^{p-1}i^{p-1}\equiv (-1)^{p-1}\equiv 1{\pmod {p}}.} {\displaystyle \prod _{i=1}^{p-1}i^{p-1}\equiv (-1)^{p-1}\equiv 1{\pmod {p}}.}

So, the truth of the Agoh–Giuga conjecture combined with Wilson's theorem would give: a number p is prime if and only if

∑ i = 1 p − 1 i p − 1 ≡ − 1 ( mod p ) {\displaystyle \sum _{i=1}^{p-1}i^{p-1}\equiv -1{\pmod {p}}} {\displaystyle \sum _{i=1}^{p-1}i^{p-1}\equiv -1{\pmod {p}}}

and

∏ i = 1 p − 1 i p − 1 ≡ 1 ( mod p ) . {\displaystyle \prod _{i=1}^{p-1}i^{p-1}\equiv 1{\pmod {p}}.} {\displaystyle \prod _{i=1}^{p-1}i^{p-1}\equiv 1{\pmod {p}}.}

See also

References