My Math Forum floor function and a constant

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 June 12th, 2012, 11:51 AM #1 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory floor function and a constant If p=2,3,5,7,11,13,17,19,... ,for which value of A; $[A^p] \text{is prime}$? do there even exists such an A? PS: [ ] sign is the floor function!
June 12th, 2012, 12:26 PM   #2
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Re: floor function and a constant

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 Originally Posted by mathbalarka do there even exists such an A?
If true, that's definitely beyond current mathematical technology to prove. If false, a disproof may be possible.

 June 12th, 2012, 09:04 PM #3 Member   Joined: May 2012 From: Chennai,India Posts: 67 Thanks: 0 Re: floor function and a constant What is A in the expression..? I do not understand. pls help.. because i see, $\left \lfloor 2.1^{19} \right \rfloor=\left \lfloor 1324849.664 \right \rfloor=1324849$ which is prime
 June 12th, 2012, 09:52 PM #4 Member   Joined: May 2012 From: Chennai,India Posts: 67 Thanks: 0 Re: floor function and a constant i am not sure what i misunderstood.. i thought all numbers $\sqrt[p_{i}]{p_{j}}$ where $p_{i}and p_{j}$ are prime numbers will be valid.
 June 12th, 2012, 10:44 PM #5 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Re: floor function and a constant In my original post, i meant that do there exist a constant, A, such that [A²], [A³], [A^5], [A^7], [A^11],... are all prime? Is it clear now?
 June 13th, 2012, 03:26 AM #6 Member   Joined: May 2012 From: Chennai,India Posts: 67 Thanks: 0 Re: floor function and a constant All prime powers should yield prime numbers... that sounds like it should be false.. i tried the brute force method, but could make only prime numbers until power of 17 or 19. For primes more than that, either the higher powers are not prime or the primes less than power of 17 or 17 become composite.. Like said by CRGreat, is it possible to prove the contradiction?
 June 13th, 2012, 03:37 AM #7 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Re: floor function and a constant looks like you dont know floor function!!! have you understood what is my problem?
June 13th, 2012, 03:46 AM   #8
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Re: floor function and a constant

Quote:
 All prime powers should yield prime numbers... that sounds like it should be false..
Nope, you haven't got it. I said for a fixed constant, floor of all prime powers of that number will yield prime number.for the definition of floor function, see, wiki,floor

 June 13th, 2012, 04:19 AM #9 Member   Joined: May 2012 From: Chennai,India Posts: 67 Thanks: 0 Re: floor function and a constant I understood the application of floor function on prime powers of A. I iterated and found that after power of 17 or 19 every number I tried (though not extensive) lead to a composite number like $\left \lfloor 1.504^p \right \rfloor for\, p\,=2,3,5,7,11 \, yields \, 2,3,7,17,89$ so i was wondering it looks like no number A can satisfy this.. so i am looking for a proof to contradict
 June 13th, 2012, 04:30 AM #10 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Re: floor function and a constant I am sure the contradiction proof will involve analytic number theory and the theory of prime gaps. Can it possibly be related with Mills' constant and the Riemann hypothesis? Sir Greathouse, we are waiting for your replies cause you know number theory better than any of us. And thanks, karthikeyan.jp for your replies.

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