In mathematics, Wolstenholme's theorem states that for a prime number p ≥ 5, the congruence
( 2 p − 1 p − 1 ) ≡ 1 ( mod p 3 ) {\displaystyle {2p-1 \choose p-1}\equiv 1{\pmod {p^{3}}}}
holds, where the parentheses denote a binomial coefficient. For example, with p = 7, this says that 1716 is one more than a multiple of 343. The theorem was first proved by Joseph Wolstenholme in 1862. In 1819, Charles Babbage showed the same congruence modulo p2, which holds for p ≥ 3. An equivalent formulation is the congruence
( a p b p ) ≡ ( a b ) ( mod p 3 ) {\displaystyle {ap \choose bp}\equiv {a \choose b}{\pmod {p^{3}}}}
for p ≥ 5, which is due to Wilhelm Ljunggren (and, in the special case b = 1, to J. W. L. Glaisher) and is inspired by Lucas's theorem. No known composite numbers satisfy Wolstenholme's theorem and it is conjectured that there are none (see below). A prime that satisfies the congruence modulo p4 is called a Wolstenholme prime (see below). As Wolstenholme himself established, his theorem can also be expressed as a pair of congruences for (generalized) harmonic numbers:
1 + 1 2 + 1 3 + ⋯ + 1 p − 1 ≡ 0 ( mod p 2 ) , and {\displaystyle 1+{1 \over 2}+{1 \over 3}+\dots +{1 \over p-1}\equiv 0{\pmod {p^{2}}}{\mbox{, and}}}
1 + 1 2 2 + 1 3 2 + ⋯ + 1 ( p − 1 ) 2 ≡ 0 ( mod p ) . {\displaystyle 1+{1 \over 2^{2}}+{1 \over 3^{2}}+\dots +{1 \over (p-1)^{2}}\equiv 0{\pmod {p}}.}
since
( 2 p − 1 p − 1 ) = ∏ 1 ≤ k ≤ p − 1 2 p − k k ≡ 1 − 2 p ∑ 1 ≤ k ≤ p − 1 1 k ( mod p 2 ) {\displaystyle {2p-1 \choose p-1}=\prod _{1\leq k\leq p-1}{\frac {2p-k}{k}}\equiv 1-2p\sum _{1\leq k\leq p-1}{\frac {1}{k}}{\pmod {p^{2}}}}
(Congruences with fractions make sense, provided that the denominators are coprime to the modulus.) For example, with p = 7, the first of these says that the numerator of 49/20 is a multiple of 49, while the second says the numerator of 5369/3600 is a multiple of 7.
Wolstenholme primes
A prime p is called a Wolstenholme prime iff the following condition holds:
( 2 p − 1 p − 1 ) ≡ 1 ( mod p 4 ) . {\displaystyle {{2p-1} \choose {p-1}}\equiv 1{\pmod {p^{4}}}.}
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