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)).
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