n ≡ 3 ( m o d 4 ) , n \equiv 3(\bmod 4), n ≡ 3 ( mod 4 ) , therefore all the prime factors of n are odd.
For contradiction, assume all of the prime factors are equivalent to 1 m o d 4. 1 \bmod 4. 1 mod 4.
Then prove n is also equivalent to 1 m o d 4 1 \bmod 4 1 mod 4 , which is a contradiction (note ( 4 k + 1 ) ( 4 m + 1 ) = 4 ( k m + k + m ) + 1 ) . (4 k+1)(4 m+1)=4(k m+k+m)+1). ( 4 k + 1 ) ( 4 m + 1 ) = 4 ( km + k + m ) + 1 ) .
Suppose for contradiction there are only finitely many primes of the form 4k+3, and call the set of all of them { p 1 , p 2 , … , p n } . \left\{p_{1}, p_{2}, \ldots, p_{n}\right\}. { p 1 , p 2 , … , p n } . But then
( p 1 p 2 ⋯ p n ) 2 + 2 = 3 ( m o d 4 ) \left(p_{1} p_{2} \cdots p_{n}\right)^{2}+2=3 \quad(\bmod 4) ( p 1 p 2 ⋯ p n ) 2 + 2 = 3 ( mod 4 )
by (1) has a prime factor p n + 1 p_{n+1} p n + 1 equivalent to 3 m o d 4 3 \bmod 4 3 mod 4 .
By definition of { p 1 , p 2 , … , p n } \left\{p_{1}, p_{2}, \ldots, p_{n}\right\} { p 1 , p 2 , … , p n } , we get p n + 1 ∈ { p 1 , p 2 , … , p n } , p_{n+1} \in\left\{p_{1}, p_{2}, \ldots, p_{n}\right\}, p n + 1 ∈ { p 1 , p 2 , … , p n } ,
Thus p n + 1 ∣ ( p 1 p 2 ⋯ p n ) 2 + 2 ⟹ p n + 1 ∣ 2 p_{n+1}\left|\left(p_{1} p_{2} \cdots p_{n}\right)^{2}+2 \Longrightarrow p_{n+1}\right| 2 p n + 1 ∣ ∣ ( p 1 p 2 ⋯ p n ) 2 + 2 ⟹ p n + 1 ∣ ∣ 2
Contradiction.
Comments