Question

Prove this is true: The sequence {4k + 3} where k ∈ N contains infinitely many primes, but the sequence {4k + 2} does not. Show your work.

Answer 1

• Below is the detailed explanation of the above mentioned problem.
• To prove : the sequence (4k+3) contains infinitely many primes, but the sequence (4k+2) does not where k belongs to set of Natural numbers.
• Prime number: A number which only has 2 factors i.e, 1 and the number itself.
• Explanation:

Let's write (4k+3) for some k, 7(k=1), 11(k=2), 15(k=3), 19(k=4), 23(k=5).................., So we can see that this sequence is an arithmetic progression(A.P) with first term as 7 and common difference as 4, and there will be many prime numbers in the sequence as we can see that for k=1,2,4,5,7,10,11,14,16,17,19,20 and so on we get a prime number in the sequence.

• Proof for (4k+3):

So to prove this above fact we use Euclid's proof of infinitude of primes, suppose there are finite primes in the sequence (4k+3) like 7(=t1), 11(=t2), 19(=t3).............(=tm),

Now let's consider a number x=(4*t₁*t₂*t₃*t₄.......t_m) -1, so this x can be rewritten as x=4((t₁*t₂*t₃*t₄.......t_m)-1)+3 which is of the form (4n+3), now as we have considered the primes in the sequence as finite so this number x of the sequence will have some prime p(of form (4n+3) ) as it's divisor but no prime from t₁*t₂*t₃*t₄.......t_m equal to p as number x is omething like (t_i*y-1) and t_i cannot divide this number so p is not equal to any of the t_i , hence there exist some p other than t₁, t₂, t₃, t₄.......t_m which divides x .Hence our assumption is wrong that there are finite prime t₁, t₂, t₃, t₄.......t_m in the sequence.  

• Proof for (4k+2):

But for sequence (4k+2), we can rewrite it as 2*(2k+1), so for every term of this sequence we have 2 as a common factor that will result in 2 as factor of that number other than 1 and the number itself, which violates the definition of a prime number.

We can also confirm it by writing the series : 6(k=1), 10(k=2), 14(k=3), 18(k=4), 22(k=5), 26(k=6) ..................(A.P with first term as 6 and common difference as 4)

So all numbers in this sequence are even numbers which are not prime.

• Hence the sequence (4k+3) contains infinite many prime and sequence (4k+2) does not contain any.

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