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natta225 [31]
3 years ago
14

Prove that there are infinitely many primes of the form 4k + 3, where k is a non-negative integer. [Hint: Suppose that there are

only finitely many such primes q1, q2, . . . , qn, and consider the number 4q1q2 · · · qn − 1.]
Mathematics
2 answers:
Mrac [35]3 years ago
7 0

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)

Simora [160]3 years ago
6 0

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.

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<em><u>Now since we want to find whose textbook reaches the ground first and by how many seconds we need to solve each equation (i.e. Eqns. (3) and (4)) at </u></em>h(t)=0<em><u>. Now since both are quadratic equations we will solve one showing the full method which can be repeated for the other one. </u></em>

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t_{1,2}=\frac{-b+/-\sqrt{b^2-4ac} }{2a} \\t_{1,2}=\frac{-60+/-\sqrt{60^2-4(-16)(40)} }{2(-16)} \\t_{1,2}=\frac{-60+/-\sqrt{6160} }{-32} \\t_{1,2}=\frac{15+/-\sqrt{385} }{8}\\\\t_{1}=4.3276 sec\\t_{2}=-0.5776 sec

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<em>Whose textbook reaches the ground first and by how many seconds?</em>

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