Let us divide the odd positive integers into two arithmetic

Let us divide the odd positive integers into two arithmetic progressions; the red numbers are 1, 5, 9, 13, 17, 21, .... The blue numbers are 3, 7, 11, 15, 19, 23, 27, .... (a) Prove that the product of two red numbers is red, and the product of two blue numbers is red. (b) Prove that a blue number has a blue prime factor. (c) Prove that there are infinitely many blue prime numbers.4^6

Solution

Red numbers are 1,5,9,13,... i.e numbers of the form 4k+1.

Blue numbers are 3,7,11,15,19,...i.e. numbers of the form 4k+3 or 4k-1.

(a) Product of two red numbers is red

r1=4n+1, r2=4m+1. r1.r2 = (4n+1)(4m+1)=16mn +4(m+n)+1 = 4k+1 which is a red number

Product of two blue numbers is red.

b1=4n+3, b2=4m+3 . b1. b2=(4n+3)(4m+3)=16mn+12(m+n)+9=4k+1 which is a red number.

(b) Any blue number has a blue prime factor.

Any blue number b=4k+3 is odd , so it will have odd prime factors only i.e. of either form 4m+1 or 4n+3.

Now let us assume b does not have any prime factors of the form 4n+3, i.e all the prime factors of b are of the form 4m+1 i.e only red prime factors.

i.e. b = (4m₁ +1)(4m₂ +1)...(4m_l +1)= 4K+1 which is again a red number .(since by result (a) product of red numbers ). This is a contradiction since b is a blue number , thus b has a blue prime factor.

Thus,any blue number has a blue prime factor.

(c) To prove: - There are infinitely many blue prime numbers.

Let us assume there are a finite number of blue prime numbers say, {p₁ ,p₂,...,p_n }

Let N =4 p₁p₂...p_n - 1. Now N is a blue number .

So, by result (b) there is a blue prime that factors N.

But each p_i divides N+1 so it cannot divide N, where i=1,2,...n.

Thus there exists a blue prime p (where p divides N) which is not equal to p_i for i=1,2,...n. which is a contradiction

Thus there are infinitely many blue primes numbers.