Proof.
Suppose that
![$ p_1, p_2, \ldots, p_n$](img239.png)
are
![$ n$](img7.png)
distinct primes.
We construct a prime
![$ p_{n+1}$](img240.png)
not equal to any of
![$ p_1,\ldots, p_n$](img241.png)
as follows. If
![$\displaystyle N=p_1 p_2 p_3 \cdots p_n + 1,$](img242.png) |
(1.2.1) |
then by Proposition
1.1.20 there is a factorization
with each
![$ q_i$](img225.png)
prime and
![$ m\geq 1$](img244.png)
.
If
![$ q_1 = p_i$](img245.png)
for some
![$ i$](img223.png)
, then
![$ p_i\mid{}N$](img246.png)
.
Because of (
1.2.1), we also have
![$ p_i\mid{}N-1$](img247.png)
, so
![$ p_i\mid{} 1=N-(N-1)$](img248.png)
, which
is a contradiction.
Thus the prime
![$ p_{n+1} = q_1$](img249.png)
is not in the list
![$ p_1,\ldots, p_n$](img241.png)
,
and we have constructed our new prime.