A natural number n is called semiprime if Omega(n)=2. My conjecture is that for any positive integer k, there are infinitely many pairs of semiprimes which differ by exactly k.
I have the following proof for your conjecture:
Let us define prime numbers p,q,r,s and positive integer k s.t. pq=rs+k, where pq and rs are semi-primes. Since k can take any positive value, we split the proof by considering when k is even, then when k is odd:
(i) Say k is even: k=2a, where a is any positive integer. We check cases when pq and rs are both even or when they are both odd (these are the only possibilities since k is even). In the first case, say pq and rs are both even and arbitrarily choose p=2, r=2 and q>=2, s>=2. Note, one can choose q=2, s=2 and p>=2, r>=2 or other possibilities. Then, pq=rs+k => 2q=2s+2a => q=s+a. Since a can take any positive integer value, the equation q=s+a holds for infinitely many values of q, s, and a. This is Polignac's conjecture for primes (http://en.wikipedia.org/wiki/Polignac%27s_conjecture). In the second case, say pq and rs are both odd. Let pq=2b+1 and rs=2c+1, where b and c are both positive integers Then, pq=rs+k => 2b+1=2c+1+2a => b=c+a. Now, we cannot assume b and c are primes (they may or may not be). However, we know that b and c can both take values from the list 4,7,10,12,16,... and are linearly related to the infinite list of semi-primes, by definition. Note, we assume an infinite list of semi-primes, since Euclid proved in 300 BC that there are infinitely many primes. Since a is arbitrary and b and c can take values from an infinite range, there are infinitely many solutions to the equation b=c+a.
(ii) Say k is odd: k=2a+1, where a is any positive integer. Then, pq is even and rs is odd or vice-versa. Let us arbitrarily choose pq to be even and rs to be odd. This means that either p=2, q=2, or p=q=2. Let p=2, and choose rs=2b-1, where b is a positive integer. This gives us pq=rs+k => 2q=2b-1+2a+1 => 2q=2b+2a => q=b+a. Like before, a is arbitrary and b is chosen from the infinite list 5,8,11,13,17,... and is related to semi-primes and, therefore, an infinite choice of primes exist. Thus, there are infinite solutions for q=b+a. QED
Tiberiu: you did not understand my question. For given k , a is not arbitrary, (in your notation) as a is either k/2 or (k-1)/2. For example, if K=6, a=3, then you are claiming there are infinitely many prime pairs differ by 3 !!!!!!! I know only one such pair (2,5). Can you please write one more for me ?
Also polignac's conjecture says that for any given even integer n, there are infinitely many prime pairs differ by n. Probably , you missed the part "for even integer n".
OK guys, thank you for your replies. Here is a better proof:
We begin by using "Dirichlet's Theorem on Primes in Arithmetic Progressions" (1837) (see, for example: http://primes.utm.edu/notes/Dirichlet.html). This theorem states that if a and b are co-prime positive integers (i.e. gcd(a,b)=1), then the sequence a, a+b, a+2b, a+3b,... contains infinitely many primes. Similarly, this sequence also contains infinitely many semi-primes (take a=1, b=1, for example, which gives us all the positive integers). Lets assume that a is fixed, b is a prime and gcd(a,b)=1 such that p=a+qb, where p and q are both primes. Then, there are infinitely many cases of the form p=a+qb. Similarly, for the semi-prime case, there are infinitely many cases for pr=a+qb, where r is a prime. Therefore, fixing a, there exist infinitely many pairs of semi-primes, pr and qb, such that pr=a+qb, where gcd(a,b)=1 and b,p,q,r are primes. QED
Tiberiu: I am very sorry to say that you have not understood Dirichlet's theorem on primes in AP. It tells us there are infinitely many primes p=a (mod b) provided a and b are coprime. It never assured that the quotient (p-a)/b is also a prime. How do you know that "q" is prime ????
Tapas: On the contrary, I understand the theorem. So, p=a mod b => p=a+qb and I have chosen q specifically such that it is prime. This is possible since q may take many values, including prime values. I suggest you study the proof in more detail. Or perhaps, you can offer your own proof if it's not too difficult. Here is a good starting place: http://www.primepuzzles.net/conjectures/conj_050.htm
Good luck!
Peter: I am sure that you have understood my problem and you know that giving an example is not equivalent to giving a proof. If some one claim that the sequence
a, a+b, a+2b, and so on contains infinitely many semiprimes for a=b=1 is equivalent to
for any (a,b)=1, the above sequence contains infinitely many semiprimes, then I am helpless. Also I know the original proof of Dirichlet for prime in AP using non-vanishing of special values of Dirichlet L-functions at 1. If my friend Tiberiu needs any references regarding Dirichlet's original proof then I can help him to give some.
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