For given positive integers a,b,c consider the set S = {a + bk + cn : k = 0, 1, 2,...,n = 0, 1, 2,...}. Characterize all triples a,b,c when S contains infinitely many primes.

Guest Mar 25, 2017

#1**+2 **__proof:__

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I'll prove that for every natural , n>1, for given positive integers a_{1}, a_{2}, .....,a_{n} and S={x_{1}a_{1}+x_{2}a_{2}+......+x_{n}a_{n}: a_{1}, a_{2}, ....,a_{n} are positive integers} S contains infinitely many primes if and only if no positive integer divides all n integers (a_{1}, a_{2},......,a_{n})

suppose there is a positive integer P, so that a_{i} is divisible by P, for every 0 1a _{1}+x _{2}a _{2}+......+x _{n}a _{n} is always divisible by P, and because there arent infinitely many primes that are divisible by any positive integer P (unless P=1, but we know P isnt 1) there arent infinitely many primes in S.

I'll prove that for every natural , n>1, for given positive integers a_{1}, a_{2}, .....,a_{n} and S={x_{1}a_{1}+x_{2}a_{2}+......+x_{n}a_{n}: a_{1}, a_{2}, ....,a_{n} are positive integers}, there exists a positive integer D, so that for every D 1, x _{2}, x _{3},......,x _{n} so that x _{1}a _{1}+x _{2}a _{2}+......+x _{n}a _{n}=y if no positive integer divides all n integers (a _{1}, a _{2},......,a _{n}).

The proof is by induction- Suppose the positive integer P divides all positive integers a_{1}, a_{2}, .....,a_{n-1}, and that after dividing each of the integers by P no other integers divides them all. Lets define b_{i}=a_{i}/P for every 0 1 1, x _{2}, x _{3},......,x _{n-1} so that x _{1}b _{1}+x _{2}b _{2}+......+x _{n-1}b _{n-1}=y Therefore, for every y that is divisible by P and bigger than P*D _{1} there exists positive integers x _{1}, x _{2}, x _{3},......,x _{n-1 }so that x _{1}a _{1}+x _{2}a _{2}+......+x _{n-1}a _{n-1}=y. Suppose a _{n} and P are coprime. Therefore, we can express every number of the form x _{1}a _{1}+x _{2}a _{2}+......+x _{n}a _{n} using the sum r _{1}*P+r _{2}*a _{n}. Using the chicken-nugget theorem (Search it up, its real) we can infer that there exists an integer D _{2} so that for every D _{2} 1 and r _{2} so that r _{1}*P+r _{2}*a _{n}=y.

(If a_{n} and P are NOT coprime that means another positive integer, F, divides both of them, and that means F divides all of the integers (a_{1}, a_{2},......,a_{n}), but that is a contradiction to what we said in the beginning- no positive integer divides all of the integers).

And therefore, if the theorem

for every natural , n>1, for given positive integers a

_{1}, a_{2}, .....,a_{n}and S={x_{1}a_{1}+x_{2}a_{2}+......+x_{n}a_{n}: a_{1}, a_{2}, ....,a_{n}are positive integers}, there exists a positive integer D, so that for every D 1, x_{2}, x_{3},......,x_{n}so that x_{1}a_{1}+x_{2}a_{2}+......+x_{n}a_{n}=y if no positive integer divides all n integers (a_{1}, a_{2},......,a_{n}).

Is true for n-1, it is also true for n.

I'll complete the theorem by proving it for n=2: suppose a_{1} and a_{2} are positive integers. if no positive integers divides both of them, then they are coprime integers, and using the chicken nugget theorem we can infer there exists a positive integer D_{1} so that for every D_{1} 1a _{1}+x _{2}a _{2}=y for some positive integers x _{1} and x _{2}.

Therefore, the theorem is true for n>1.

We know there are infinitely many primes, and we know that for every positive integer D there exists a FINITE number of primes between 1 and D, So there are infinitely many primes that are equal or bigger than D. from the following theorem, we can infer Every prime that is bigger than D will be in the set S, and therefore, there are infinitely many primes in S IF AND ONLY IF no positive integer divides a_{1}, a_{2},......,a_{n}.

Now, this was just a generalization of your question, and I'll use it to answer your question-

Suppose b and c are coprime. Therefore, there exists a positive integer D so that for every D 1b+x _{2}c=y for some positive integers x _{1} and x _{2}. therefore, for every a+D 1b+x _{2}c=y for some positive integers x _{1} and x _{2}. That means there are infinitely many primes in S.

so if b and c are coprime, we can infer that there are infinitely many primes in S. We also know that if there exists a positive integer P so that it divides a, b and c There is a finite number of primes in S.

The question is: what if there exists a positive integer P so that P divides b, and c, but not a? that is equivalent to the question: Is there an arithmetic sequence a_{n}=a+bn where a and b are coprime and that contains a finite number of primes?

unfortunately i cant answer that question, BUT i have this cool generalization (?) of the question, and i love generalizations, so i'll keep it although its useless.

EDIT: The answer to the question *"Is there an arithmetic sequence a _{n}=a+bn where a and b are coprime and that contains a finite number of primes?" *is NO, and therefore S contains infinitely many primes only and only if no positive integer divides both a, b, and c. But We have to prove it using Dirichlet's theorem on arithmetic progressions, So i cant really prove it.

Who gave you this question? because whoever did must be a horrible person.

Ehrlich
Jun 24, 2017