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Topic: Twain levels, the third level and prime numbers

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Re: Twain levels, the third level and prime numbers

Hello, notree, you wrote:  a word "least" before x?

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Re: Twain levels, the third level and prime numbers

The first counter x, starting to sort out x in the order of 1,2,3,4,5,6...

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Re: Twain levels, the third level and prime numbers

Hello, notree, you wrote: N> the first counter x, starting to sort out x in the order of 1,2,3,4,5,6... Prove that such x generally is.

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Re: Twain levels, the third level and prime numbers

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Re: Twain levels, the third level and prime numbers

Unfortunately prove itself I can not, therefore asked about the help. Convinced me of it a considerable quantity of the examples confirming the statement.

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Re: Twain levels, the third level and prime numbers

Hello, T4r4sB, you wrote: TB> Prove that such x generally is. "For any simple p, there is the least x such that the least divider 2^x-x^3 is equal p" At x=p+8 expression 2^x-x^3 always shares on p [2 ^ (p+8) - (p+8) ^3] %p = 0 [2 ^ (p+8) = (p+8) ^3] %p [2 ^ (p+8-p+1) = 8^3] %p [2^9 = 2^9] %p And time is though any x that is and the least. Over remaining it is necessary to think in the afternoon.

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Re: Twain levels, the third level and prime numbers

Hello, kov_serg, you wrote: _> At x=p+8 expression 2^x-x^3 always shares on p About, cool! And now make so that it did not share on any smaller p.

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Re: Twain levels, the third level and prime numbers

Hello, T4r4sB, you wrote: TB> Hello, kov_serg, you wrote: _>> At x=p+8 expression 2^x-x^3 always shares on p TB> About, cool! And now make so that it did not share on any smaller p. For any simple p is x1=p+2, x2=p+8, x3=2p+4 which always do expression 2^x-x^3 by multiple p

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Re: Twain levels, the third level and prime numbers

Hello, kov_serg, you wrote: _> For any simple p is x1=p+2, x2=p+8, x3=2p+4 which always do expression 2^x-x^3 by multiple p On a condition still it is necessary, that it did not share on any smaller prime number.

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Re: Twain levels, the third level and prime numbers

Hello, T4r4sB, you wrote: TB> Hello, kov_serg, you wrote: _>> At x=p+8 expression 2^x-x^3 always shares on p TB> About, cool! And now make so that it did not share on any smaller p. Your statement is not true Here an example p=29 x=31 2^x-x^3 = 2^31-31^3 = 2147453857 = 23*29*59*197*277 It shares on 23

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Re: Twain levels, the third level and prime numbers

Hello, T4r4sB, you wrote: TB> Hello, kov_serg, you wrote: _>> For any simple p is x1=p+2, x2=p+8, x3=2p+4 which always do expression 2^x-x^3 by multiple p TB> On a condition still it is necessary, that it did not share on any smaller prime number. For this purpose it would be necessary that p-6 not a prime number.

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Re: Twain levels, the third level and prime numbers

Hello, kov_serg, you wrote: _> your statement not truly Which statement? And it why your answer has no relation to the citation. Tell how to receive number for which f (x) shares on p, but does not share on any q smaller p.

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Re: Twain levels, the third level and prime numbers

Hello, kov_serg, you wrote: _> For this purpose it would be necessary that p-6 not a prime number. That is you want to tell, what for p=11 the answer is not present?

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Re: Twain levels, the third level and prime numbers

Hello, T4r4sB, you wrote: TB> Hello, kov_serg, you wrote: _>> For this purpose it would be necessary that p-6 not a prime number. TB> that is you want to tell, what for p=11 the answer is not present? I want to tell that p=11 x=p+2=13 2^13-13^3 = 5995 and it shares on 5, and 5 it is less than 11 still separately a special case for 3 p=3 x=2*p+4=10 2^x-x^3=24 but it is not interesting

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Re: Twain levels, the third level and prime numbers

Hello, kov_serg, you wrote: _> I want to tell that _> p=11 x=p+2=13 _> 2^13-13^3 = 5995 and it shares on 5, and 5 it is less than 11 Yes, your decision means does not allow to find the answer for p=11.

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Re: Twain levels, the third level and prime numbers

Hello, T4r4sB, you wrote: TB> Hello, kov_serg, you wrote: _>> your statement is not true TB> Which statement? The statement that p the least factor. TB> and it why your answer has no relation to the citation. Tell how to receive number for which f (x) shares on p, but does not share on any q smaller p. If is_prime (p-6) =0 then at f (p+2) minimum  p

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Re: Twain levels, the third level and prime numbers

Hello, T4r4sB, you wrote: TB> Hello, kov_serg, you wrote: _>> I want to tell that _>> p=11 x=p+2=13 _>> 2^13-13^3 = 5995 and it shares on 5, and 5 it is less than 11 TB> Yes, your decision means does not allow to find the answer for p=11. And here my decision for p=11 is 3 decisions x=13,19,26 all these decisions do f (x) the multiple 11 least 13 gives 5995, it shares on 11 but also on 5 which less than 11. I.e.  the statement directed by the task false. To remove the contradiction it is possible demanding is_prime (p-6) =0 then all a bundle.

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Re: Twain levels, the third level and prime numbers

Hello, kov_serg, you wrote: _> the Statement that p the least factor. Not I stated it, this statement of the problem, read attentively! _> if is_prime (p-6) =0 then at f (p+2) minimum  p And if is not present?

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Re: Twain levels, the third level and prime numbers

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Re: Twain levels, the third level and prime numbers

Hello, kov_serg, you wrote: _> And here my decision for p=11 is 3 decisions x=13,19,26 all these decisions do f (x) by multiple 11 _> the least 13 gives 5995, it shares on 11 but also on 5 which less than 11. _> i.e.  the statement directed by the task false. Read a condition attentively. It is required to prove that exists such x that f (x) has a minimum divider p. You now refuted the statement that "minimum x such that f (x) shares on p, cannot share on smaller numbers". Yes, it is incorrect, but in a condition not it is told, read a condition attentively.

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Re: Twain levels, the third level and prime numbers

Hello, T4r4sB, you wrote: TB> Hello, kov_serg, you wrote: _>> the Statement that p the least factor. TB> not I stated it, this statement of the problem, read attentively! _>> if is_prime (p-6) =0 then at f (p+2) minimum  p TB> And if is not present? Then precisely there is a divider p-6 which too idle time is_prime (p-6) =1

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Re: Twain levels, the third level and prime numbers

Hello, T4r4sB, you wrote: TB> Hello, kov_serg, you wrote: _>> And here my decision for p=11 is 3 decisions x=13,19,26 all these decisions do f (x) multiple 11 _>> the least 13 gives 5995, it shares on 11 but also on 5 which less than 11. _>> i.e.  the statement directed by the task false. TB> read a condition attentively. It is required to prove that exists such x that f (x) has a minimum divider p. You now refuted the statement that "minimum x such that f (x) shares on p, cannot share on smaller numbers". Yes, it is incorrect, but in a condition not it is told, read a condition attentively. Affirms that  x1=p+2, x2=p+8, x3=2*p+6 which do so f (x) shares on p. Others are not present. - here an error found, there are also others For example p=31 x=14 f (x) =2^3*5*11*31 p=239 x=11 f (x) =3*239 p=937 x=36 f (x) =2^6*5*7*29*937*1129

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Re: Twain levels, the third level and prime numbers

Hello, kov_serg, you wrote: _> Affirms that  x1=p+2, x2=p+8, x3=2*p+6 which do so f (x) shares on p. Others are not present. The incorrect statement. There are also others x such that f (x) shares on p.

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Re: Twain levels, the third level and prime numbers

Hello, T4r4sB, you wrote: TB> Hello, kov_serg, you wrote: _>> Affirms that  x1=p+2, x2=p+8, x3=2*p+6 which do so f (x) shares on p. Others are not present. TB> the Incorrect statement. There are also others x such that f (x) shares on p. Here you are right, forgot one more level of freedom.