Prove that there are infinitely many primes of the form 4k+34k + 3.

Answer: There are infinitely many primes of the form 4k+34k + 3.

Step-by-step solution

Step 1: Assume a finite number of primes

To prove that there are infinitely many primes of the form 4k+34k+3, we use proof by contradiction. We start by assuming that there is a finite number of such primes. Let this finite set of primes be P={p1,p2,…,pn}P = \{p_1, p_2, \dots, p_n\}, where each pip_i is of the form 4k+34k+3.

Step 2: Construct a new number N

Now, let's construct a new number NN. We define NN as four times the product of all these primes, minus one. This number NN can also be written in the form 4k+34k+3, since 4(p1p2…pn)4(p_1 p_2 \dots p_n) is a multiple of 4, and subtracting 1 gives a number of the form 4k+34k+3.

Step 3: Analyze the prime factors of N

Every prime factor of NN must be of the form 4k+14k+1 or 4k+34k+3. If all prime factors of NN were of the form 4k+14k+1, then their product NN would also be of the form 4k+14k+1. However, NN is of the form 4k+34k+3. Therefore, NN must have at least one prime factor of the form 4k+34k+3.

Step 4: Contradiction

Let pjp_j be a prime factor of NN that is of the form 4k+34k+3. According to our initial assumption, pjp_j must be one of the primes in the set P={p1,p2,…,pn}P = \{p_1, p_2, \dots, p_n\}. This means pjp_j divides the product p1p2…pnp_1 p_2 \dots p_n. If pjp_j divides NN and pjp_j divides p1p2…pnp_1 p_2 \dots p_n, then pjp_j must also divide their difference: N−4(p1p2…pn)=−1N - 4(p_1 p_2 \dots p_n) = -1. This implies pjp_j divides −1-1, which is impossible for a prime number.

Step 5: Conclusion

Since our assumption leads to a contradiction, the initial assumption that there is a finite number of primes of the form 4k+34k+3 must be false. Therefore, there are infinitely many primes of the form 4k+34k+3.

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