I'll prove that for every natural , n>1, for given positive integers a1, a2, .....,an and S={x1a1+x2a2+......+xnan: a1, a2, ....,an are positive integers} S contains infinitely many primes if and only if no positive integer divides all n integers (a1, a2,......,an)
proof:
suppose there is a positive integer P, so that ai is divisible by P, for every 0 1a 1+x 2a 2+......+x na 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 a1, a2, .....,an and S={x1a1+x2a2+......+xnan: a1, a2, ....,an are positive integers}, there exists a positive integer D, so that for every D 1, x 2, x 3,......,x n so that x 1a 1+x 2a 2+......+x na n=y if no positive integer divides all n integers (a 1, a 2,......,a n).
proof:
The proof is by induction- Suppose the positive integer P divides all positive integers a1, a2, .....,an-1, and that after dividing each of the integers by P no other integers divides them all. Lets define bi=ai/P for every 0 1 1, x 2, x 3,......,x n-1 so that x 1b 1+x 2b 2+......+x n-1b 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 1a 1+x 2a 2+......+x n-1a n-1=y. Suppose a n and P are coprime. Therefore, we can express every number of the form x 1a 1+x 2a 2+......+x na 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 an and P are NOT coprime that means another positive integer, F, divides both of them, and that means F divides all of the integers (a1, a2,......,an), 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 a1, a2, .....,an and S={x1a1+x2a2+......+xnan: a1, a2, ....,an are positive integers}, there exists a positive integer D, so that for every D 1, x 2, x 3,......,x n so that x 1a 1+x 2a 2+......+x na 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 a1 and a2 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 D1 so that for every D1 1a 1+x 2a 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 a1, a2,......,an.
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 2c=y for some positive integers x 1 and x 2. therefore, for every a+D 1b+x 2c=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 an=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 an=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.
