## euler's polynomial proof

without using a counter example, show a proof that Euler's polynomial equation P(n)=n^2+n+41 can not be used to generate all the primes.can a general proof be made to show that all primes cannot be generated by a specific polynomial?
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 Recognitions: Gold Member Science Advisor Staff Emeritus Well, what, exactly, is "Euler's polynomial equation"? Don't you think that is important?
 thats the proposition made by Euler that P(n)=n^2+n+41 can generate all primes and it was suddenly proven false by using a counter example, isnt there any way he can be proven to be false?

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## euler's polynomial proof

 Quote by khotsofalang thats the proposition made by Euler that P(n)=n^2+n+41 can generate all primes and it was suddenly proven false by using a counter example, isnt there any way he can be proven to be false?
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.
 Blog Entries: 2 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, but as shown by this link that was false. The proof is simple an noted in the link. Assume that P(x) is a polynominal in x that gives a prime $$p$$ for the value $$n$$. But then $$p|P(n+kp)$$ for all integer $$k$$ 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?

It is possible with polynomials in more variables, as the same wiki article points out:

 ...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).
 Recognitions: Homework Help Science Advisor That doesn't count - look at the restriction placed on both the input and output variables.

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 Quote by Count Iblis It is possible with polynomials in more variables, as the same wiki article points out:
but the same polynominal would generate composites also with large enough variables. Think how easy it would be to exceed the largest prime otherwise
 P(n) = n^2 + n + 41 => P(40) = 40^2 + 40 + 41 = 40 * 40 + 40 + 41 = 40 * 41 + 41 = 41 * 41 = composite.

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 Quote by ramsey2879 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?
 Recognitions: Gold Member That no polynominal can generate all primes is easily shown using the reference given already http://en.wikipedia.org/wiki/Formula_for_primes Let P(1) = a prime S. $$Then P(1+kS) \equiv P(1) \equiv 0 Mod S.$$

 Quote by ramsey2879 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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 Quote by robert Ihnot That no polynominal can generate all primes is easily shown using the reference given already http://en.wikipedia.org/wiki/Formula_for_primes Let P(1) = a prime S. $$Then P(1+kS) \equiv P(1) \equiv 0 Mod S.$$