Homepage › Solution manuals › Ivan Niven › An Introduction to the Theory of Numbers › Exercise 2.7.11* ($\sigma_{p-2} \equiv p \sigma_{p-3} \pmod {p^3}.$)

An Introduction to the Theory of Numbers

Exercise 2.7.11* ($\sigma_{p-2} \equiv p \sigma_{p-3} \pmod {p^3}.$)

Let p be a prime, p ≥ 5 , and suppose that the numbers σ j are as in (2.7). Show that σ p − 2 ≡ p σ p − 3 ( mod p 3 ) .

Answers

Proof. The σ i are defined by

f ( x ) = ( x − 1 ) ( x − 2 ) ⋯ ( x − p + 1 ) = x p − 1 − σ 1 x p − 2 + σ 2 x p − 3 − ⋯ − σ p − 2 x + σ p − 1 ,

thus

f ( p ) = ( p − 1 ) ! = p p − 1 − σ 1 p p − 2 + σ 2 p p − 3 − ⋯ + σ p − 3 p 2 − σ p − 2 p + σ p − 1 ,

Since ( p − 1 ) ! = σ p − 1 , on subtracting this amount from both sides and dividing through by p , we deduce that

p p − 2 − σ 1 p p − 3 + ⋯ − σ p − 4 p 2 + σ p − 3 p − σ p − 2 = 0 . ( see p. 96 )

We know that σ j ≡ 0 ( mod p ) for 1 ≤ j ≤ p − 2 (see p. 96). Since p ≥ 5 , σ p − 4 ≡ 0 ( mod p ) and all the others terms except the two last are divisible by p 3 . Therefore σ p − 3 p − σ p − 2 ≡ 0 ( mod p 3 ) , so

σ p − 2 ≡ p σ p − 3 ( mod p 3 ) .

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2024-09-04 16:03
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