Exercise 6, § 2
Functions of a Real Variable, VI, p. 24
a) The rational numbers being defined by formula (20) of VI, p. 7, denote by the denominator of written in the form of an irreducible fraction. Now prove that no prime factor of can be (use the recurrence formula (23) of VI, p. 8). b) Now prove that one has, for every integer and every integer
c) Deduce from b) by induction on that, for every prime number the denominator of written in the form of an irreducible fraction, is not divisible by (observe that cannot divide ). d) Deduce from c) that the number
where runs through the set of prime numbers and is even, is an integer. Conclude that
where runs through the set of prime numbers such that divides , is an integer (Clausen-von Staudt theorem; use Exercise 5 e).
Solution
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