Proving Prime Numbers: The Equation Test for Primality

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Discussion Overview

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.

Discussion Character

  • Exploratory
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant asks how to prove that if \( p \) is prime, then a derived equation of \( p \) is also prime.
  • Another participant clarifies the inquiry by asking for specifics about what is meant by "a derived equation of \( p \)."
  • It is suggested that while \( (2^p - 1) \) can be prime if \( p \) is prime, it is not universally true, as shown with \( p = 11 \).
  • A participant questions how to prove that a polynomial expressed in terms of \( p \) always yields a prime number when \( p \) is prime, and whether such a polynomial exists.
  • One participant asserts that it has been proven that such polynomials do not exist, citing that if \( P(n) \) is prime, then \( P(n + a \cdot \text{prime}) \) is also divisible by "prime."
  • Another participant mentions that there exists a function that produces primes but is considered "totally useless."
  • Further clarification is requested regarding the divisibility argument presented about polynomials and primes.

Areas of Agreement / Disagreement

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.

Contextual Notes

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.

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?
 
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Hi l-1j-cho! :smile:

What exactly do you mean with "a derived equation of p"?
 
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
 
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
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?
 
Last edited:
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)
 
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)
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.
 
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thank you!
 
ramsey2879 said:
P(n+a*prime) is also divisible by prime,
QUOTE]

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.

The fun thing that there IS such a function! The function is totally useless, but it exists: http://www.math.hmc.edu/funfacts/ffiles/10003.5.shtml
 
  • #10
l-1j-cho said:
ramsey2879 said:
P(n+a*prime) is also divisible by prime,
QUOTE]

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.
 

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