Advanced book on Mathematics Olympiad

5.2 Arithmetic 255

an example, in this topology both the set of odd integers and the set of even integers are
open. Because the intersection of two arithmetic progressions is an arithmetic progres-
sion, the open sets ofTare precisely the unions of arithmetic progressions. In particular,
any open set is either infinite or void.
If we define

Aa,d={...,a− 2 d,a−d,a,a+d,a+ 2 d,...},a∈Z,d> 0 ,

thenAa,dis open by hypothesis, but it is also closed because it is the complement of the
open setAa+ 1 ,d∪Aa+ 2 ,d∪···∪Aa+d− 1 ,d. HenceZ\Aa,dis open.
Now let us assume that only finitely many primes exist, sayp 1 ,p 2 ,...,pn. Then

A 0 ,p 1 ∪A 0 ,p 2 ∪···∪A 0 ,pn=Z\{− 1 , 1 }.

This union of open sets is the complement of the open set

(Z\A 0 ,p 1 )∪(Z\A 0 ,p 2 )∪···∪(Z\A 0 ,pn);

hence it is closed. The complement of this closed set, namely{− 1 , 1 }, must therefore
be open. We have reached a contradiction because this set is neither empty nor infinite.
Hence our assumption was false, and so there are infinitely many primes. 

Let us begin with the examples.

Example.Prove that for all positive integersn, the number

33

n

+ 1

is the product of at least 2n+1 not necessarily distinct primes.

Solution.We induct onn. The statement is clearly true ifn=1. Because

33

n+ 1

+ 1 =

(

33

n

+ 1

)(

32 ·^3

n

− 33

n

+ 1

)

,

it suffices to prove that 3^2 ·^3
n
− 33
n
+1 is composite for alln≥1. But this follows from
the fact that

32 ·^3

n

− 33

n

+ 1 =

(

33

n

+ 1

) 2

− 3 · 33

n

=

(

33

n

+ 1

) 2

−

(

3

3 n 2 + 1 )^2

is the product of two integers greater than 1, namely,

33

n

+ 1 − 3

3 n 2 + 1

and 3^3

n

+ 1 − 3

3 n 2 + 1

.

This completes the induction.