n≡3(mod4), therefore all the prime factors of n are odd.
For contradiction, assume all of the prime factors are equivalent to 1mod4.
Then prove n is also equivalent to 1mod4 , which is a contradiction (note (4k+1)(4m+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 {p1,p2,…,pn}. But then
(p1p2⋯pn)2+2=3(mod4)
by (1) has a prime factor pn+1 equivalent to 3mod4 .
By definition of {p1,p2,…,pn} , we get pn+1∈{p1,p2,…,pn},
Thus pn+1∣∣(p1p2⋯pn)2+2⟹pn+1∣∣2
Contradiction.
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