Amod Posted November 24, 2007 Share Posted November 24, 2007 The little I know about primes is that there is no any 'Systematic Formula' for obtaining prime numbers in either ascending or descending order.And the whole world is in a delima as wheter there is a limit to primes or NOT.That is whetther there is a HIGHEST PRIME NUMBER. I have Good, Better and Best formulae for generating Prime Numbers systematically. TRY THE GOOD ONE: P=7d-4 Where: P=prime number(Needed) d=odd number(Chosen) Example: At d=5, it means; P=7(5)-4=35-4=31 there are a few exceptions for the Good formula. Questions and contributions are welcome Link to comment Share on other sites More sharing options...

insane_alien Posted November 24, 2007 Share Posted November 24, 2007 there are a few exceptions for the Good formula. which essentially means you have not come up with a formula that systematically finds primes and could be used in finding a limit or even if there is one. actually, for this, every second number it produces is even. it also does not include ALL prime numbers. but please, post the other formulas i will make a spreadsheet to compare them and post it here. Link to comment Share on other sites More sharing options...

uncool Posted November 24, 2007 Share Posted November 24, 2007 which essentially means you have not come up with a formula that systematically finds primes and could be used in finding a limit or even if there is one. actually, for this, every second number it produces is even. it also does not include ALL prime numbers. but please, post the other formulas i will make a spreadsheet to compare them and post it here. A few things: First, to insane_alien: he said d must be odd, which gives the formula 14d - 11. Now, to Amod: There are an infinite number of exceptions to this formula. Any d divisible by 11 will create a number divisible by 11, so the formula doesn't work. However, there is a theorem called Dirischlet's theorem that says that an infinite number of outputs from your formula will be prime. Another thing to Amod: It is known that there are an infinite number of primes. In fact, it's been known for over 2000 years (by Euclid) that any finite list of primes is incomplete. Here's the proof: Assume there are a finite number of primes. Let them be in the set S. Then take the product of all the elements of S, and add 1. Let this number be A. A must have a prime divisor. However, this prime divisor cannot be in S - so there is a prime number not on the list, so the list is incomplete. This means that the number of primes must be infinite. Now, as a note: there is also a simple proof that any polynomial cannot always generate prime numbers. Do you want it? =Uncool- Link to comment Share on other sites More sharing options...

JaKiri Posted November 24, 2007 Share Posted November 24, 2007 The little I know about primes is that there is no any 'Systematic Formula' for obtaining prime numbers in either ascending or descending order.And the whole world is in a delima as wheter there is a limit to primes or NOT.That is whetther there is a HIGHEST PRIME NUMBER.I have Good, Better and Best formulae for generating Prime Numbers systematically. TRY THE GOOD ONE: P=7d-4 Where: P=prime number(Needed) d=odd number(Chosen) Example: At d=5, it means; P=7(5)-4=35-4=31 there are a few exceptions for the Good formula. Questions and contributions are welcome What exactly is this supposed to do? Link to comment Share on other sites More sharing options...

the tree Posted November 24, 2007 Share Posted November 24, 2007 Now, as a note: there is also a simple proof that any polynomial cannot always generate prime numbers. Do you want it?That sounds pretty interesting. Link to comment Share on other sites More sharing options...

uncool Posted November 24, 2007 Share Posted November 24, 2007 That sounds pretty interesting. Alright; it basically goes like this: Assume you have a polynomial such that for all n >= 0 (n is an integer), f(n) is prime. Then f(0) is prime. Let it be called p. Then f(a*p) must be divisible by p, as polynomials are preserved modulo a prime. Therefore, f(a*p) must be p for all a, as the only prime that p divides is itself. However, a polynomial can only be a value infinitely many times if it is constant. Therefore, any polynomial that is always prime is constant - and therefore trivial. =Uncool- Link to comment Share on other sites More sharing options...

Country Boy Posted November 25, 2007 Share Posted November 25, 2007 And the whole world is in a delima as wheter there is a limit to primes or NOT.That is whetther there is a HIGHEST PRIME NUMBER. Not since Euclid, a couple thousand years ago! Suppose there were a largest prime number. p. Multiply all the prime numbers up to p together and add 1. That number is not divisible by any prime number less than or equal to p since you would have a remainder of 1. That means that either this new number is prime itself or there is some other prime number, larger than p. Either way, there cannot be a largest prime number. Link to comment Share on other sites More sharing options...

insane_alien Posted November 29, 2007 Share Posted November 29, 2007 okay, so i missed the part about the odd numbers. still, every d ending in 7 is divisible by 5 and there are a bunch of other non-primes in there too. not to mention is comes nowhere near listing all the primes. so, where are these 'better' formulae? Link to comment Share on other sites More sharing options...

D H Posted November 29, 2007 Share Posted November 29, 2007 Jones, Wada, Sato, and Weins, "Diophantine representation of the set of prime numbers", Amer. Math. Monthly 83(1976), 449-464 Abstract: In this paper it is proved that the set of prime numbers is exactly the set of positive values of a polynomial of the 25th degree' date=' in 26 variables as the variables range over the nonnegative integers: (k+2){1 – [noparse'][[/noparse]wz+h+j–q]^{2} – [(gk+2g+k+1)(h+j)+h–z]^{2} – [2n+p+q+z–e]^{2} – [16(k+1)^{3}(k+2)(n+1)^{2}+1–f^{2}]^{2} – [e^{3}(e+2)(a+1)^{2}+1–o^{2}]^{2} – [(a^{2}–1)y^{2}+1–x^{2}]^{2} – [16r^{2}y^{4}(a^{2}–1)+1–u^{2}]^{2} – [((a+u^{2}(u^{2}–a))^{2} –1)(n+4dy)^{2} + 1 – (x+cu)^{2}]^{2} – [n+l+v–y]^{2} – [(a^{2}–1)l^{2}+1–m^{2}]^{2} – [ai+k+1–l–i]^{2} – [p+l(a–n–1)+b(2an+2a–n^{2}–2n–2)–m]^{2} – [q+y(a–p–1)+s(2ap+2a–p^{2}–2p–2)–x]^{2} – [z+pl(a–p)+t(2ap–p^{2}–1)–pm]^{2}} Each positive value of the above polynomial is prime. Also each prime is a value of the above polynomial, for some nonnegative integers a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s,t,u,v,w,x,y,z. Thus the set of prime numbers is identical with the set of positive values of the polynomial. Link to comment Share on other sites More sharing options...

insane_alien Posted November 29, 2007 Share Posted November 29, 2007 i was referring to the ones amod has 'discovered' Link to comment Share on other sites More sharing options...

D H Posted November 29, 2007 Share Posted November 29, 2007 i was referring to the ones amod has 'discovered' I'm sorry. I thought you might want to add a polynomial that actually does find primes to your spreadsheet. I see now that your spreadsheet is reserved for polynomials that purport to find primes, but fail to achieve that objective. Now, as a note: there is also a simple proof that any polynomial cannot always generate prime numbers. Do you want it?=Uncool- The polynomial f(x)=3 generates a prime for all x. 1 Link to comment Share on other sites More sharing options...

doG Posted November 29, 2007 Share Posted November 29, 2007 And the whole world is in a delima as wheter there is a limit to primes or NOT.That is whetther there is a HIGHEST PRIME NUMBER. There's no dilemma. In an infinite number set there should be an infinite number of primes. Look at how a prime sieve works and it should be obvious that an infinite number set will always have numbers as you ascend that will not be a factor of any previous primes in the sieve. Link to comment Share on other sites More sharing options...

uncool Posted November 29, 2007 Share Posted November 29, 2007 I'm sorry. I thought you might want to add a polynomial that actually does find primes to your spreadsheet. I see now that your spreadsheet is reserved for polynomials that purport to find primes, but fail to achieve that objective. The polynomial f(x)=3 generates a prime for all x. Gah...bad terminology. I meant to say that it wouldn't work for any nontrivial (nonconstant). =Uncool- Link to comment Share on other sites More sharing options...

D H Posted November 29, 2007 Share Posted November 29, 2007 Gah...bad terminology. I meant to say that it wouldn't work for any nontrivial (nonconstant).=Uncool- Just toying with ya. A trivial polynomial isn't much of much value. I don't think Jone's polynomial has much more than academic interest, either It was an interesting offshoot of the work done on Hilbert's tenth problem. On the other hand, the polynomial-time AKS primality testing algorithm (for-the-masses review and the paper itself) won immediate accolades and awards, including a nomination for the Fields medal. Link to comment Share on other sites More sharing options...

uncool Posted November 29, 2007 Share Posted November 29, 2007 Just toying with ya. A trivial polynomial isn't much of much value. I don't think Jone's polynomial has much more than academic interest, either It was an interesting offshoot of the work done on Hilbert's tenth problem. On the other hand, the polynomial-time AKS primality testing algorithm (for-the-masses review and the paper itself) won immediate accolades and awards, including a nomination for the Fields medal. Yeah...now that's an interesting find - that prime testing can be done in poly time of the log of the prime. What is the current least prime-finding algorithm so far? =Uncool- Link to comment Share on other sites More sharing options...

D H Posted November 29, 2007 Share Posted November 29, 2007 Yeah...now that's an interesting find - that prime testing can be done in poly time of the log of the prime. What is the current least prime-finding algorithm so far?=Uncool- Do you mean factoring? That is an NP problem, and finding a way to solve that in polynomial time is worth big $$$. Link to comment Share on other sites More sharing options...

uncool Posted November 29, 2007 Share Posted November 29, 2007 Do you mean factoring? That is an NP problem, and finding a way to solve that in polynomial time is worth big $$$. Nah, I mean actually finding a prime. I know the usual methods today are simply finding a pseudoprime, which is very likely to be prime, by checking fermat's little theorem. However, that doesn't guarantee the primality of the number found. =Uncool- Link to comment Share on other sites More sharing options...

Malteaser Posted June 25, 2009 Share Posted June 25, 2009 Was having a bit of fun with Fermat's last theorem and, when factorising x cubed minus y cubed, I came up with an interesting formula: x^2 - xy +y^2. I was looking at this formula to see if it would factorise, since the solution appears complex, I concluded that it didn't. However, I then realised that it's more important to solve x^2 - xy +y^2 + N. Which does have roots, I think, depending on the value of N. When I tried to solve it I came accross the function 3/4x^2, which can only be prime if x is an even number. i quickly realised this was a trivial statement. Bit it made me think that my original equation must be a prime if x and y are odd and mutually exclusive (though I'm not sure why I have to say the mutually exclusive bit). 1. Can someone help me to understand whether this is true and if not which numebrs x and y make it not true and why this is the case? 2. Also can some direct me to the polynomial theorem discussed above and whether this covers "polynomials" with an x and a y in them - are they still described as polynomials? 3. What is the actual question then re: prime numbers. I understood that the question was that there are no known formula that will always create a prime number. Is that the right question. Thanks Merged post follows: Consecutive posts mergedHaving done a simple spreadsheet my statement "it made me think that my original equation must be a prime if x and y are odd and mutually exclusive (though I'm not sure why I have to say the mutually exclusive bit)." is not true, But I'd still like to work out why it's not true if I can't find roots to my equation? and some tips on which way tio investigate this further? Link to comment Share on other sites More sharing options...

Weejonnie Posted August 20, 2009 Share Posted August 20, 2009 Originally Posted by Jones et al In this paper it is proved that the set of prime numbers is exactly the set of positive values of a polynomial of the 25th degree, in 26 variables as the variables range over the nonnegative integers: (k+2){1 – [wz+h+j–q]2 – [(gk+2g+k+1)(h+j)+h–z]2 – [2n+p+q+z–e]2 – [16(k+1)3(k+2)(n+1)2+1–f2]2 – [e3(e+2)(a+1)2+1–o2]2 – [(a2–1)y2+1–x2]2 – [16r2y4(a2–1)+1–u2]2 – [((a+u2(u2–a))2 –1)(n+4dy)2 + 1 – (x+cu)2]2 – [n+l+v–y]2 – [(a2–1)l2+1–m2]2 – [ai+k+1–l–i]2 – [p+l(a–n–1)+b(2an+2a–n2–2n–2)–m]2 – [q+y(a–p–1)+s(2ap+2a–p2–2p–2)–x]2 – [z+pl(a–p)+t(2ap–p2–1)–pm]2} Each positive value of the above polynomial is prime. Also each prime is a value of the above polynomial, for some nonnegative integers a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s,t,u,v,w,x,y, z. Thus the set of prime numbers is identical with the set of positive values of the polynomial. ------------------------------------------------------------------------------ I may be missing something obvious here but surely this function is the result of multiplying two numbers (k+2) and {} together and hence can't be prime unless (k+2) equals 1 (or {} equals 1.) If we set k+2 = 1 then k=-1 and a cursory examination of the second (larger) section of the formula shows that in most cases 'k' appears close to a '+1) Although the formula is quoted by de sautoy (music of the primes) this fact (+ the incredible coincidence that the number of variables 'just happens' to be the same as the number of characters in the alpahabet) suggests someone is playing a practical joke. Can someone enlighten me? Link to comment Share on other sites More sharing options...

D H Posted August 20, 2009 Share Posted August 20, 2009 It is the number k+2 that is prime, not the product. If you can find an integer k and some set of non-negative integers a, b, ..., z such that the value of the polynomial is positive then k+2 is a prime number. If k+2 is a prime number, there always exists some set of non-negative integers a, b, ..., z that yield a positive value. In fact, the term in the brackets will be one in this case. The polynomial will evaluate to k+2. In the case that k+2 is not prime, there is no set a, b, ..., z that yields a positive value for the polynomial. Link to comment Share on other sites More sharing options...

John Cuthber Posted August 20, 2009 Share Posted August 20, 2009 I didn't understand most of that but I think this remains both relevant and ammusing. http://xkcd.com/622/ Link to comment Share on other sites More sharing options...

WarriorClass III Posted August 16, 2011 Share Posted August 16, 2011 Here is a solution for obtaining any prime: https://sites.google.com/site/primenumbertheory/home/the-prime-thesis The program to run it is also included. Link to comment Share on other sites More sharing options...

khaled Posted August 16, 2011 Share Posted August 16, 2011 If I remember right, some mathematicians such as Mersenne have a formula to generate a large prime number, Mersenne Prime: [math]M_p = 2^p - 1[/math] Link to comment Share on other sites More sharing options...

uncool Posted August 16, 2011 Share Posted August 16, 2011 If I remember right, some mathematicians such as Mersenne have a formula to generate a large prime number, Mersenne Prime: [math]M_p = 2^p - 1[/math] It works pretty often, but not always. You still have to check that the output is prime. =Uncool- Link to comment Share on other sites More sharing options...

imatfaal Posted August 17, 2011 Share Posted August 17, 2011 It works pretty often, but not always. You still have to check that the output is prime. =Uncool- Yep - you do have to check still. There are some very good checking algorithms - which I believe are less intensive for potential mersenne primes than for other candidates. All the really big primes are mersenne. If you have spare cpu cycles that you are willing to share - http://www.mersenne.org/ Link to comment Share on other sites More sharing options...

## Recommended Posts

## Create an account or sign in to comment

You need to be a member in order to leave a comment

## Create an account

Sign up for a new account in our community. It's easy!

Register a new account## Sign in

Already have an account? Sign in here.

Sign In Now