5. Squarefree and smooth numbers 9

Proof. We use not only that ζ(2) =

π2/6

but also that ζ(4) =

π4/90.

(Again see Exercise 5.) Thus

ζ(2)2/ζ(4)

= 5/2. The Euler factorization

(1.2) implies that

5

2

=

ζ(2)2

ζ(4)

=

p

(1 −

p−4)(1

−

p−2)−2

=

p

p4

− 1

p4

p4

(p2 − 1)2

=

p

p2

+ 1

p2 − 1

,

so that

5

2

=

5

3

·

10

8

·

26

24

· · · .

If there are only finitely many primes, then the product on the right-hand

side is a finite one and can be written as M/N, where M = 5 · 10 · 26 · · ·

and N = 3 · 8 · 24 · · · . Then M/N = 5/2, so 2M = 5N. Since 3 | N, it must

be that 3 | M. But this cannot be: M is a product of numbers of the form

k2

+ 1, and no such number is a multiple of 3.

Wagstaff has asked whether one can give a more elementary proof that

5/2 =

p

p2+1

p2−1

. The discussion of this (open) question in [Guy04, B48] was

the motivation for the preceding proof of Theorem 1.1.

5. Squarefree and smooth numbers

Recall that a natural number n is said to be squarefree if it is not divisible

by the square of any integer larger than 1. The fundamental theorem of

arithmetic shows that there is a bijection

{finite subsets of the primes} ←→ {squarefree positive integers},

given by sending

S −→

p∈S

p.

So to prove the infinitude of the primes, it suﬃces to prove that there are

infinitely many positive squarefree integers.

J. Perott’s proof, 1881. We sieve out the non-squarefree integers from

1,...,N by removing those divisible by

22,

then those divisible by

32,

etc.

The number of removed integers is bounded above by

∞

k=2

N/k2

≤ N

∞

k=2

k−2

= N(ζ(2) − 1),

so that the number of squarefree integers up to N, say A(N), satisfies

(1.6) A(N) ≥ N − N(ζ(2) − 1) = N(2 − ζ(2)).