**Author** |
**Message** |

**Rajesh Bhowmick** Veteran poster
Post Number: 708
| Posted on Tuesday, 28 February, 2012 - 08:31 pm: | |
"Any number excluding 1,6,8,9 & the prime numbers can be expressed as the sum of its prime factors & certain prime numbers provided there should be no repetition of the prime numbers in the summation in any way."For e.g. 12=2*2*3 12=2+3+7 Is there anymore counterexample to it? regards. |

**lebesgue** Veteran poster
Post Number: 2700
| Posted on Tuesday, 28 February, 2012 - 08:42 pm: | |
What about 22? |

**Rajesh Bhowmick** Veteran poster
Post Number: 709
| Posted on Tuesday, 28 February, 2012 - 08:57 pm: | |
lebesgue, thanx for adding another one in the sequence. any other one? Regards. |

**Rajesh Bhowmick** Veteran poster
Post Number: 710
| Posted on Tuesday, 28 February, 2012 - 09:25 pm: | |
Include 4 in the list also. regards. |

**lebesgue** Veteran poster
Post Number: 2701
| Posted on Tuesday, 28 February, 2012 - 09:46 pm: | |
It looks better if you view any prime as the sum of its prime factors (with empty set of additional primes needed); so with minor change in your point of view you can have result (strictly speaking, conjecture, at the moment) with finitely many exceptions only. |

**lebesgue** Veteran poster
Post Number: 2702
| Posted on Tuesday, 28 February, 2012 - 09:58 pm: | |
Rajesh's conjecture: Any integer n>22 can be expressed as the sum of distinct prime numbers such that all prime factors of n are among them. Another version of this conjecture: The above is true for any positive integer n except 1,4,6,8,9,22. |

**Rajesh Bhowmick** Veteran poster
Post Number: 713
| Posted on Wednesday, 29 February, 2012 - 05:56 am: | |
If the above is true then these are also true 1)If n be any even integer & the summation of its prime factors is odd then the odd number remaining would be a prime number &/or if not prime then if the above exceptions are subtracted from it then with minimum two exceptions that remaining odd number would give a prime number. 2)If n be any odd integer & the summation of its prime factors is even then the odd number remaining would be a prime number &/or if not prime then if the above exceptions are subtracted from it then with minimum two exceptions that remaining odd number would give a prime number. regards. |

**lebesgue** Veteran poster
Post Number: 2704
| Posted on Wednesday, 29 February, 2012 - 09:14 am: | |
The versions that I formulated above without restrictions on number of primes you are allowed to add after adding prime factors of n, are much simpler than those with such restrictions. With restrictions it becomes the Coldbach's conjecture applied to N=n-(sum of prime factors of n) and even more difficult as you have then restrictions on primes used to add up to N (you are not allowed to use factors of n again). |

**lebesgue** Veteran poster
Post Number: 2705
| Posted on Wednesday, 29 February, 2012 - 09:41 am: | |
As I mentioned in our previous discussion in another thread it is very easy to prove (using Bertrand's postulate, say) that every integer n>6 can be written as a sum of distinct primes. Your conjectures (without restrictions on the number of summands we can add after adding prime factors of n) are similar, but more technically complicated, as for given n you need to represent N=n-(sum of prime factors of n) as sum of distinct primes, with additional complication that we can't use any prime factor of n again. But this is doable, and I would be able to write simple proof that Rajesh's conjecture is true for all sufficiently large n. If you can't find more small counterexamples, then version provided by me are apparently true. But versions with Goldbach's conjecture flavor I wouldn't be able to attack, unfortunately, it is much more difficult. |

**Rajesh Bhowmick** Veteran poster
Post Number: 714
| Posted on Wednesday, 29 February, 2012 - 11:08 am: | |
lebesgue, If Goldbach's conjecture is not true then how can mine be & ultimately the Bertrands theorem? O.K let me try to be more clear. Pls refer to my post 713. If in that statement the remaining number is even then it has to be expressible as difference of min. two primes & max. it can be any number of primes in the summation according to my statement (keeping all the conditions of my statement). The only even numbers u need to check manually are the even numbers from the exceptions i.e. 4,6,8 & 22. Except these all must have min two primes as its summations. now according to Goldbach's conjecture if we check these four numbers they are satisfying Goldbach's conjecture. 4=2+2 6=3+3 8=5+3=2+3+3=2+2+2+2 22=19+3.=13+3+3+3=11+2+3+3+3=8+3+3+3+3+2=5+3+3+3+3+3+2.=11+11 According to Goldbach's conjecture.Now if my statement is correct & & restate Goldbach's conjecture for Primes to be "distinct Primes" then only 4 & 6 are the only exceptions in that statement. regards. |

**lebesgue** Veteran poster
Post Number: 2707
| Posted on Wednesday, 29 February, 2012 - 11:31 am: | |
It seems to be perfectly consistent to admit that Goldbach's conjecture is untrue but your conjectures are true (which I believe I could prove relatively easily, at least for large n). I can't follow your arguments in your previous post, I am afraid, unless you state precisely what you are assuming (like assuming Goldbach's conjecture true,... or assuming strenghtened Goldbach's conjecture: 'any larger even n is the sum of two distinct primes' true...), and what you are trying to conclude from that. |

**Rajesh Bhowmick** Veteran poster
Post Number: 715
| Posted on Wednesday, 29 February, 2012 - 12:42 pm: | |
O.K. lebesgue,Let me be more clear. If I restate Goldbach's conjecture like this that "Any even number greater then 6 can be expressed as summation of "at least" two distinct prime numbers." Now If my statement is true as u have described then this statement is also true. As u see If u take any even number, then 2 would be one of its prime factor. So the summation of the prime factors would contain 2 & this summation would be either even or odd. Now I am talking abt the even case. Then the other number would also be an even number & according to my given statement in this post, this "other" even number(except 4 & 6) should be expressible as sum of two distinct prime numbers other wise how would then my statement be true if we are not able to express this other even number as sum of two distinct primes as in my statement which u claim to have proof (for any "n" except those of the exceptions) there should be primes & that also distinct & added some of the primes(including 2) would have been used up in the summation of the factors. So the only conclusion is this that the statement made by me in this post is true for any even n>6. The min. number u have to check is now 8 then & I have shown it.The proof for larger "n" u have so proving my modified Goldbach's conjecture of this post side by side. regards. |

**lebesgue** Veteran poster
Post Number: 2708
| Posted on Wednesday, 29 February, 2012 - 01:42 pm: | |
The statement: "Any integer n>6 is the sum of distinct primes" is much easier than what I called above Rajesh's conjecture; it can be based on Bertrand's postulate and induction only. It is something on national maths olympiad level,I would say, so you are invited to start with this one to make progress. Rajesh's conjecture is above imo level, but still doable. Goldbach's conjecture is certainly of different league. |

**Rajesh Bhowmick** Veteran poster
Post Number: 719
| Posted on Wednesday, 29 February, 2012 - 01:53 pm: | |
"Any integer n>6 is the sum of distinct primes" Correction "Any integer n>6 is the sum of 'at least two' distinct primes". regards. |

**lebesgue** Veteran poster
Post Number: 2709
| Posted on Wednesday, 29 February, 2012 - 02:01 pm: | |
I don't quite agree - usually we allow one prime, and then n>6 is correct. For even n(>2) we needn't add 'at least two' because even n can't be just one prime. But some small primes like 11 can't be sum of at least two distinct primes, I believe, so the bound n>something should be higher then. |

**Rajesh Bhowmick** Veteran poster
Post Number: 720
| Posted on Wednesday, 29 February, 2012 - 02:06 pm: | |
Correction "Any even integer n>6 is the sum of 'at least two' distinct primes". I just copied ur statement & I didn't noticed the "even" part. regards. |

**Rajesh Bhowmick** Veteran poster
Post Number: 721
| Posted on Wednesday, 29 February, 2012 - 02:12 pm: | |
11 itself is a prime. why to express it as sum of two distinct primes then? I'm afraid I really didn't got u now lebesgue. regards. |

**lebesgue** Veteran poster
Post Number: 2710
| Posted on Wednesday, 29 February, 2012 - 03:36 pm: | |
I am happy with sum of just ONE prime, you came forward with the requirement 'sum of at least two', in post 719, for example. But it is still true that sufficiently large integers are sum of at least two distinct primes, you just have to increase the bound n>6 slightly higher; 11 and 17 are not of such form. |

**Rajesh Bhowmick** Veteran poster
Post Number: 723
| Posted on Wednesday, 29 February, 2012 - 05:12 pm: | |
lebesgue, I'm talking abt n being "even" when I say n>6.(refer to post 720 of mine) I think u r talking abt "n" being any positive integer including the odd numbers which I have not stated yet. But If u want to take my view & want to state the theorem for any "n" then exclude all prime numbers including 4 & 6.U will surely get ur results. regards. |

**Rajesh Bhowmick** Veteran poster
Post Number: 725
| Posted on Wednesday, 29 February, 2012 - 06:52 pm: | |
U can state that for "certain" prime numbers also. regards. |

**Rajesh Bhowmick** Veteran poster
Post Number: 742
| Posted on Thursday, 05 April, 2012 - 08:01 am: | |
More ideas or views are welcome. regards. |