Respuesta :
Answer:
The prove is as given below
Step-by-step explanation:
Suppose there are only finitely many primes of the form 4k + 3, say {p1, . . . , pk}. Let P denote their product.
Suppose k is even. Then P ≅ 3^k (mod 4) = 9^k/2 (mod 4) = 1 (mod 4).
ThenP + 2 ≅3 (mod 4), has to have a prime factor of the form 4k + 3. But pₓ≠P + 2 for all 1 ≤ i ≤ k as pₓ| P and pₓ≠2. This is a contradiction.
Suppose k is odd. Then P ≅ 3^k (mod 4) = 9^k/2 (mod 4) = 1 (mod 4).
Then P + 4 ≅3 (mod 4), has to have a prime factor of the form 4k + 3. But pₓ≠P + 4 for all 1 ≤ i ≤ k as pₓ| P and pₓ≠4. This is a contradiction.
So this indicates that there are infinite prime numbers of the form 4k+3.
Answer:
From the explanation below, the number of primes of the form 4k+3 cannot be finite and if that be the case, the opposite is true that there are infinitely many primes of the form (4k +3)
Step-by-step explanation:
Let q1,q2,…,qn be odd primes of the form 4k+3.
We can write their products as P= (q1xq2....... qr) for some r integer; (4q1+3)x(4q2+3)x..... (4qr + 3)
Let's consider the number N, where
N=4q1q2…qn-1.
It is clear that none of the qi divides N, and that 4 does not divide N.
Since N is odd and greater than 1, it is a product of one or more odd primes.
Now, we'll show that at least one of these primes is of the form 4k+3.
The prime divisors of N cannot be all of the shape 4k+1 because the product of any number of not necessarily distinct primes of the form 4k+1 is itself of the form 4k+1.
But N is not of the form 4k+1. So some prime p of the form 4k+3 divides N.
We have already seen that p cannot be one of q,…,qn.
Thus, it follows that given any collection {q1,…,qn} of primes of the form 4k+3, there is a prime p of the same form which is not in the collection.
Thus, the number of primes of the form 4k+3 cannot be finite and if that be the case, the opposite is true that there are infinitely many primes of the form (4k +3)
