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- Thread starter khotsofalang
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- #2

HallsofIvy

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Well, what, exactly, **is** "Euler's polynomial equation"? Don't you think that is important?

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- #4

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I believe you mean that P(n) would be prime for positive integers n, but there is now a general proof out there that no polynominal equation can generate only primes for all positive integers. I am sure that someone can cive you a link to a proof.

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But then [tex]p|P(n+kp)[/tex] for all integer [tex]k[/tex] so these numbers must be composite for every k so the polynominal is a constant, p, rather than a polynominal.

P.S. Euler never said that the polynominal generated "all primes", but then that is obviously not what you meant is it?

- #6

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...is a polynomial inequality in 26 variables, and the set of prime numbers is identical to the set of positive values taken on by this polynomial inequality as the variables a, b, …, z range over the nonnegative integers.

A general theorem of Matiyasevich says that if a set is defined by a system of Diophantine equations, it can also be defined by a system of Diophantine equations in only 9 variables. Hence, there is a prime-generating polynomial as above with only 10 variables. However, its degree is large (in the order of 10^45). On the other hand, there also exists such a set of equations of degree only 4, but in 58 variables (Jones 1982).

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matt grime

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That doesn't count - look at the restriction placed on both the input and output variables.

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but the same polynominal would generate composites also with large enough variables. Think how easy it would be to exceed the largest prime otherwiseIt is possible with polynomials in more variables, as the same wiki article points out:

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P(40) = 40^2 + 40 + 41

= 40 * 40 + 40 + 41

= 40 * 41 + 41 = 41 * 41 = composite.

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CRGreathouse

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See also http://en.wikipedia.org/wiki/Formula_for_primes Euler's formula for primes as posted in this thread was once widely thought to generate primes for any integer n

Really?

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Let P(1) = a prime S. [tex] Then P(1+kS) \equiv P(1) \equiv 0 Mod S.[/tex]

- #12

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but the same polynominal would generate composites also with large enough variables. Think how easy it would be to exceed the largest prime otherwise

The positive values of the Jones polynomial for positive values of the variables are always primes and all primes can be obtained this way. The problem is just that the probability that a random choice of the input vaiables will yield a positive value is astronomically small. If finally, after billions of years of trying, some positive value is obtained, it will likely be a prime number like 23.

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CRGreathouse

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Let P(1) = a prime S. [tex] Then P(1+kS) \equiv P(1) \equiv 0 Mod S.[/tex]

Was that in response to my post? I was incredulous about the claim that "once widely thought to generate primes for any integer n". It's obvious that no nonconstant polynomial can produce primes for all n, so it's hard for me to believe that this was ever believed. But it could be true, so I was looking for ramsey2879 or someone else to comment on that.

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I am 60 and some years old so I some times don't remember exactly what I read too well. The book is up in the attic. Some time I will look again at what was said about the polynominal. But that the polynominal does give composite values is of course easily shown.Was that in response to my post? I was incredulous about the claim that "once widely thought to generate primes for any integer n". It's obvious that no nonconstant polynomial can produce primes for all n, so it's hard for me to believe that this was ever believed. But it could be true, so I was looking for ramsey2879 or someone else to comment on that.

- #15

CRGreathouse

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If you do, I'd appreciate the effort if you post it.

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