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?
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.