l-1j-cho
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how do you prove, if p is prime, then a derived equation of p is prime, if true?
The discussion revolves around the concept of proving whether certain derived equations or polynomials yield prime numbers when evaluated at prime inputs. Participants explore the relationship between prime numbers and polynomial expressions, questioning the existence of such polynomials that consistently produce primes.
Participants express differing views on the existence of polynomials that yield primes for all prime inputs. While some assert that such polynomials do not exist, others suggest that there are functions that can produce primes under certain conditions, leading to an unresolved debate.
The discussion includes assumptions about the nature of polynomials and their behavior with respect to prime numbers, as well as references to specific mathematical properties that remain unverified within the conversation.
You can't be saying "if p is prime, (2^p - 1) is prime" because that is not true for p = 11. But it is easy to show that (2^p -1) can be prime only if p is prime. As for equations being prime, do you mean proving that a polynominal is not factorable into the product of two polynominals of lower degree? Or do you mean to show that a certain polynomial in n yields primes for all positive values of n less than a certain integer?l-1j-cho said:hello!
uhm, I don't know anything in particular, but something like
if p is prime, the following equation is prime
or if p is prime, (2^p -1) is prime such things
It has been proven that such polynominals don't exist since if P(n) is prime then P(n+a*prime) is also divisible by "prime",l-1j-cho said:oh right my apology
I mean polynomials that is expressed in terms of p.
obviously, polynomials like p^2+5p+6 is not prime because if can be factored to (p+2)(p+3)
but my question is, how do you prove that a random polynomials always spits out a prime number whenever we plug in a prime number?
but before that, would such polynomial exist?
(not necesarrily polynomials but exponents or other stuff)
ramsey2879 said:P(n+a*prime) is also divisible by prime,
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sorry but, could you explain about this more?
ramsey2879 said:It has been proven that such polynominals don't exist since if P(n) is prime then P(n+a*prime) is also divisible by "prime",
. Don't know but very much doubt that there is some exotic function in P that is prime for all prime P.
l-1j-cho said:ramsey2879 said:P(n+a*prime) is also divisible by prime,
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sorry but, could you explain about this more?
Sure 3p +2 = 23 for p = 7, so for p = 7 + a*23 e.g. for p = 7,30,53,... ,3p + 2 is divisible by 23 and hence not prime. The same goes for any polynominal in p.