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 2012-05-04, 12:54 #1 Lee Yiyuan   Feb 2011 Singapore 5×7 Posts Binomial Primes [P.S. Sorry if i posted in the wrong section, and sorry for my bad English too] Please note that prime numbers here excludes 1 and includes 2. Firstly, allow me to introduce and define some terms. Binomial numbers : Numbers of the form a^n +/- b^n, where a,b,n are integers >1 Decremental binomial numbers : Binomial numbers of the form (a+1)^n -a^n Decremental binomial primes : Decremental binomial numbers which are prime Binomial exponent : The value of n in a decremental binomial number of the form (a+1)^n -a^n, where a,n are integers >1 Binomial base : The value of a in a decremental binomial number of the form (a+1)^n -a^n, where a,n are integers >1 I have came out with the following statements, and the tried to prove/disprove them. Statement 1 : All primes of the form a^n - b^n, where a,b,n are natural numbers, must be decremental primes. Statement 2 : All decremental binomial primes must have a prime binomial exponent. Statement 3 : All decremental binomial primes must have a prime binomial base. Statement 4 : There exists infinitely many decremental binomial primes. Statement 5 : There exists a deterministic test to determine the primality of a decremental binomial number for all positive binomial exponents and binomial base. So far, I have only managed to prove statements 1 and 2, and disprove statement 3. (Although i am unsure of the validity of my proofs) Proof of statement 1 : Let a^n - b^n be a prime, where a,b,n are integers >1. By polynomial factorization, a^n - b^n = (a - b)[a^(n-1) + a^(n-2)b +... + ab^(n-2) + b^(n-1)] It follows that a - b is a factor of a^n - b^n. Since a^n - b^n is a prime, a - b = +/-(a^n - b^n) or +/- 1 The only solution to a - b = +/-(a^n - b^n) is a = b. For a = b, a^n - b^n = 0, hence a - b = +/-1 For a - b = -1, a = b - 1. For a = b - 1, a^n - b^n <= 0. Therefore, a = b + 1. It follows that for a binomial number of the form a^n - b^n to be prime, it must be a decremental binomial number. QED Proof of statement 2 : Let (a+1)^n - a^n be a prime, where a,n are natural numbers. Suppose n is composite. It can then be expressed in the form n = r X s for some r and s. By binomial expansion, (a+1)^n - a^n = (a+1)^(rs) - a^(rs) =[(a+1)^r - a^r][a^(r(s-1)) + a^(r(s-2))b +... + b^(s(r-1))] Hence, if n is composite, (a+1)^n - a^n will always be factorized into 2 distinct numbers, making it composite, contradicting it being prime. It follows that the binomial exponent must be prime. QED (Dis)Proof of statement 3: A counter example is 7^2 - 6^2 = 13 Here, the binomial base is 6, which is composite. However, i have yet to proof or disprove statements 4 and 5, and i am here today to kindly ask you for your help in helping me with improving the statements and their respective proofs/disproofs, and also to proof/disproof statements 4 and 5. If possible, it would be great to come out with a deterministic test for decremental binomial numbers and a proof of correctness for this test as well. Last fiddled with by fivemack on 2012-05-04 at 14:19
2012-05-04, 13:07   #2
science_man_88

"Forget I exist"
Jul 2009
Dartmouth NS

2×3×23×61 Posts

Quote:
 Originally Posted by Lee Yiyuan [P.S. Sorry if i posted in the wrong section, and sorry for my bad English too] Firstly, allow me to introduce and define some terms. Binomial numbers : Numbers of the form a^n +/- b^n Decremental binomial numbers : Binomial numbers of the form (a+1)^n -a^n Binomial primes : Decremental binomial numbers which are prime Binomial exponent : The value of n in a decremental binomial number of the form (a+1)^n -a^n Binomial base : The value of a in a decremental binomial number of the form (a+1)^n -a^n I have came out with the following statements, and the tried to prove/disprove them. Statement 1 : For a binomial number to be prime, it must be a decremental binomial number. Statement 2 : For a decremental binomial number to be prime, it's binomial exponent must be prime. Statement 3 : For a decremental binomial number to be prime, it's binomial base must be prime. Statement 4 : There exists infinitely many binomial primes. Statement 5 : There exists a deterministic test to determine the primality of a decremental binomial number. So far, I have only managed to prove statements 1 and 2, and disprove statement 3. (Although i am unsure of the validity of my proofs) Proof of statement 1 : Let a^n - b^n be a prime. By binomial factorization, a^n - b^n = (a - b)[a^(n-1) + a^(n-2)b +... + ab^(n-2) + b^(n-1)] It follows that a - b is a factor of a^n - b^n. Since a^n - b^n is a prime, a - b = +/-(a^n - b^n) or +/- 1 The only solution to a - b = +/-(a^n - b^n) is a = b. For a = b, a^n - b^n = 0, hence a - b = +/-1 For a - b = -1, a = b - 1. For a = b - 1, a^n - b^n <= 0. Therefore, a = b + 1. It follows that for a binomial number to be prime, it must be of the form (a+1)^n - a^n. QED Proof of statement 2 : Let (a+1)^n - a^n be a prime. Suppose n is composite. It can then be expressed in the form n = r X s for some r and s. By binomial expansion, (a+1)^n - a^n = a^(rs) - a^(rs) =[(a+1)^r - a^r][a^(r(s-1)) + a^(r(s-2))b +... + b^(s(r-1))] Hence, if n is composite, (a+1)^n - a^n will always be factorized into 2 distinct numbers, making it composite, contradicting it being prime. It follows that the binomial exponent must be prime. QED (Dis)Proof of statement 3: A counter example is 7^2 - 6^2 = 13 Here, the binomial base is 6, which is composite. However, i have yet to proof or disprove statements 4 and 5, and i am here today to kindly ask you for your help in helping me with improving the statements and their respective proofs/disproofs, and also to proof/disproof statements 4 and 5. If possible, it would be great to come out with a deterministic test for decremental binomial numbers and a proof of correctness for this test as well.
a=2; 3^2+2^2 -> 9+4->13 a prime

2012-05-04, 13:13   #3
Lee Yiyuan

Feb 2011
Singapore

2316 Posts

Quote:
 Originally Posted by science_man_88 a=2; 3^2+2^2 -> 9+4->13 a prime
True, but what i meant by binomial primes were numbers of the form (a+1)^n - a^n
This of course is my own definition of a binomial prime (i couldnt find any definitions after a google search on the term)

2012-05-04, 13:16   #4
science_man_88

"Forget I exist"
Jul 2009
Dartmouth NS

2·3·23·61 Posts

Quote:
 Originally Posted by Lee Yiyuan True, but what i meant by binomial primes were numbers of the form (a+1)^n - a^n This of course is my own definition of a binomial prime (i couldnt find any definitions after a google search on the term)
Quote:
 For a binomial number to be prime, it must be a decremental binomial number.
you define "a binomial number" as a^n$\pm$b^n so it's possible to have 3^2+2^2 as a binomial number and it's prime and it's not a decremental binomial number under your definitions. this equality disproves your first statement.

Last fiddled with by science_man_88 on 2012-05-04 at 13:17

2012-05-04, 13:18   #5
Lee Yiyuan

Feb 2011
Singapore

2316 Posts

Quote:
 Originally Posted by science_man_88 you define "a binomial number" as a^n$\pm$b^n so it's possible to have 3^2+2^2 as a binomial number and it's prime and it's not a decremental binomial number under your definitions. this equality disproves your first statement.
Sorry sir, i missed out a decremental before binomial prime in statement 1. Fixed it now.

2012-05-04, 13:26   #6
science_man_88

"Forget I exist"
Jul 2009
Dartmouth NS

2·3·23·61 Posts

Quote:
 Originally Posted by Lee Yiyuan Sorry sir, i missed out a decremental before binomial prime in statement 1. Fixed it now.
ran a PARI/gp script this is also false if you have no limit on n :

10^1-3^1 -> 7 -> prime 10-3 =7 not 1.

2012-05-04, 13:32   #7
Lee Yiyuan

Feb 2011
Singapore

5·7 Posts

Quote:
 Originally Posted by science_man_88 ran a PARI/gp script this is also false if you have no limit on n : 10^1-3^1 -> 7 -> prime 10-3 =7 not 1.

Thank you for pointing out, ive now limited n to integers more than one in the definition of a binomial number.

Last fiddled with by Lee Yiyuan on 2012-05-04 at 13:33

2012-05-04, 13:41   #8
science_man_88

"Forget I exist"
Jul 2009
Dartmouth NS

841810 Posts

Quote:
 Binomial primes : Decremental binomial numbers which are prime
Quote:
 Statement 1 : For a binomial number of the form a^n - b^n to be prime, it must be a decremental binomial number.
these 2 clearly show the same thing, namely, that you only intend that decremental binomials can be prime.

of course:

OEIS seq A138389 :
Quote:
 Binomial primes: positive integers n such that every i not coprime to n and not exceeding n/2 does not divide binomial(n-i-1,i-1).
though I feel like I forgot to leave criticism for R.D.Silverman to make.

Last fiddled with by science_man_88 on 2012-05-04 at 13:44

2012-05-04, 13:48   #9
Lee Yiyuan

Feb 2011
Singapore

5×7 Posts

Quote:
 Originally Posted by science_man_88 these 2 clearly show the same thing, namely, that you only intend that decremental binomials can be prime. of course: OEIS seq A138389 : though I feel like I forgot to leave criticism for R.D.Silverman to make.

Fixed once again. I am indeed in debt for your guidiance.

2012-05-04, 13:50   #10
science_man_88

"Forget I exist"
Jul 2009
Dartmouth NS

841810 Posts

Quote:
 Originally Posted by Lee Yiyuan Fixed once again. I am indeed in debt for your guidiance.
sad part is I'm one that needs the guidance half the time.

2012-05-04, 14:00   #11
Lee Yiyuan

Feb 2011
Singapore

5·7 Posts

Quote:
 Originally Posted by science_man_88 sad part is I'm one that needs the guidance half the time.
But you have determination, and i respect you for that!
On a related note, have you any idea how to contact the admins? I seem to have troubles editing my thread post as i need to change "natural numbers" to "integers more than 1"

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