# Thread: A bit different from Goldbach's conjecture.

1. Captain Kremmen, U have seen the wrong thread in ask nrich it is actually "Is it known" of Ask nrich site not "Are these facts known" thread of ask nrich site?
regards.

2. Originally Posted by indianmath
He actually got it confused with Goldbach's conjecture not Beal's formula.

I can't see how it's a reformulation of Golbach's conjecture.
If he wants to say that, he needs to show how you derive one from the other.

"Any odd number which is not a prime, except the number nine, is the sum of all its prime factors, plus other primes.
In this summation no number is used more than once."

3. Originally Posted by Captain Kremmen
Thanks for that.
Here.
https://nrich.maths.org/discus/messa...tml?1330266813

Lebesgue, on that site, seems to think it is a reformulation of Beal's Conjecture..
Opinions here? Is it?
It seems to me that is is NOT a reformulation of that conjecture, but I will leave it to the mathematicians.

This is Beal's Conjecture.

BEAL'S CONJECTURE: If Ax + By = Cz, where A, B, C, x, y and z are positive integers and x, y and z are all greater than 2, then A, B and C must have a common prime factor.
See http://www.bealconjecture.com/

Could someone explain to me, simply, what Fermats last theorem has to do with Prime numbers?

@Indianmath.
Do you always need all the prime factors, or can some occasionally be left out?
You need to get the formula right or someone will pinch your idea.
I told u in the summation primes should not be get repeated .for e.g.
25=5*5
25=5+17+3(I have taken 5 once since 5 can't be repeated in the summation)
regards.

4. I hope this isn't of use to codebreakers, otherwise you'll have the CIA or its equivalent knocking on your door.

5. Originally Posted by Captain Kremmen
I can't see how it's a reformulation of Golbach's conjecture.
If he wants to say that, he needs to show how you derive one from the other.

"Any odd number which is not a prime, except the number nine, is the sum of all its prime factors, plus other primes.
In this summation no number is used more than once."
"In this summation no prime number is used more than once."
Just add the red portion in ur statement. It would make a more clearer view then.
regards.

6. Originally Posted by Captain Kremmen
I hope this isn't of use to codebreakers, otherwise you'll have the CIA or its equivalent knocking on your door.
Captain Kremmen, How can this be used by codebreakers? Are u serious abt ur comment? I simply made a statement which is true & totally a mathematical work.
regards.

7. indianmath, your comments about being able to prove this are ambiguous, particularly at the NRICH link. Can you prove this mathematically or is it a pure conjecture at this point?

8. Originally Posted by indianmath
Captain Kremmen, How can this be used by codebreakers? Are u serious abt ur comment? I simply made a statement which is true & totally a mathematical work.
regards.
Yes I am joking. But on the other hand, primes are used for making codes. It might help people to find primes if they can quickly discard non primes.
It could have uses in cryptography, but possibly to make codes rather than break them.

9. Originally Posted by RJBeery
indianmath, your comments about being able to prove this are ambiguous, particularly at the NRICH link. Can you prove this mathematically or is it a pure conjecture at this point?
RJBeery, If Bertrand's theorem is true then my theorem is also true as u see after 9 the next number which comes into my theorem's context is 15 which i have shown in the statement & if u want to extend it more i.e for any number then except for 2,6 & 9 u will find no exception of my theorem.
regards.

10. Originally Posted by rpenner
Specifically, though, it seems he wants to include the prime factors of the number in the sum.

We can define a prime factorization of a number as:
$\forall n \in \mathbb{N} \quad \exists F_n \; \left( F_n \subset V \times \mathbb{P} \; \wedge \; n = \prod_{k \in \textrm{dom}\, F_n} F_n(k) \right)$

From this we can describe two different functions on n
$f(n) = n - \sum_{k \in \textrm{dom}\, F_n} F_n(k) \\ g(n) = n - \sum_{k \in \textrm{range} \, F_n} k$

For 12, f(12) = 12 - 2 - 2 - 3 = 5 but g(12) = 12 - 2 - 3 = 7. I don't think the OP distinguished these cases enough.

So as written, the statement is:
$\forall k \in \mathbb{N} \setminus \left{ 4 \right} \; 2k+1 \not\in \mathbb{P} \; \rightarrow \; \exists A_{\tiny 2k+1} \subset \mathbb{P} \setminus ( \text{range} \, F_{\tiny 2k+1} ) \; \left( 2k+1 = \sum_{i \in \textrm{dom}\, F_{\tiny 2k +1}} F_{\tiny 2k +1}(i) + \sum_{j \in A_{\tiny 2k +1}} j \right)$
"For every odd, natural number other than 9, if the number is not prime, there is subset of primes not part of the prime factorization the number that when totaled and added to the prime factors (including multiplicity) of the number equal that number."

But 1 is not the sum of any primes.
And the primes trivially satisfy this with $A_{p} = \emptyset$, so I don't see why the odd primes are treated differently than other odd numbers.
Finally, I don't see why odd numbers are treated differently than even numbers.

$n \in \mathbb{N} \; \wedge \; n \gt 18 \; \rightarrow \; \exists A_{n} \subset \mathbb{P} \setminus ( \text{range} \, F_{n} ) \; \left( n = \sum_{i \in \textrm{dom}\, F_{n}} F_{n}(i) + \sum_{j \in A_{n}} j \right)$ seems like a less convoluted way to express the same idea.

I think 1, 6, 9, 18 are the only natural numbers that can't be partitioned into their prime factors (including multiplicity) and possibly additional unique primes not common to their prime factorization, but obviously I haven't invested any time looking.
18=2*3*3
18=2+3+13(As there should not be any repetition in the summation so only one 3 has to be taken from the factors.)
regards.

11. Ah, that was the piece I needed.

Shorter, stronger claim:
"All natural numbers other than 1, 4, 6, 8, 9, or 22 can be partitioned into unique primes such that all of their prime factors are present in the partition."
1 can't be partitioned into primes
Any prime can only be partitioned one way into primes including its factors, i.e., into itself
4, 6, 8, 9, 22 can't be partitioned into unique primes once their prime factors are accounted for.
Here is an (updated) list of how some small composite numbers could be partitioned:
10 = ( 2 + 5 ) + ( 3 )
12 = ( 2 + 3 ) + ( 7 )
14 = ( 2 + 7 ) + ( 5 )
15 = ( 3 + 5 ) + ( 7 )
16 = ( 2 ) + ( 3 + 11 )
18 = ( 2 + 3 ) + ( 13 )
20 = ( 2 + 5 ) + ( 13 )
21 = ( 3 + 7 ) + ( 11 )
24 = ( 2 + 3 ) + ( 19 )
25 = ( 5 ) + ( 3 + 17 )
26 = ( 2 + 13 ) + ( 11 )
27 = ( 3 ) + ( 5 + 19 )
28 = ( 2 + 7 ) + ( 19 )
30 = ( 2 + 3 + 5 ) + ( 7 + 13 )

A more interesting question is, for each natural number, how many distinct partitions into unique primes that include all prime factors there are. Note that zero can be partitioned into the empty set, so it has more solutions than 1.

12. Originally Posted by rpenner
Ah, that was the piece I needed.

Shorter, stronger claim:
"All natural numbers other than 1, 4, 6, 8, 9, or 22 can be partitioned into unique primes such that all of their prime factors are present in the partition."
1 can't be partitioned into primes
Any prime can only be partitioned one way into primes including its factors, i.e., into itself
4, 6, 8, 9, 22 can't be partitioned into unique primes once their prime factors are accounted for.
Here is an (updated) list of how some small composite numbers could be partitioned:
10 = ( 2 + 5 ) + ( 3 )
12 = ( 2 + 3 ) + ( 7 )
14 = ( 2 + 7 ) + ( 5 )
15 = ( 3 + 5 ) + ( 7 )
16 = ( 2 ) + ( 3 + 11 )
18 = ( 2 + 3 ) + ( 13 )
20 = ( 2 + 5 ) + ( 13 )
21 = ( 3 + 7 ) + ( 11 )
24 = ( 2 + 3 ) + ( 19 )
25 = ( 5 ) + ( 3 + 17 )
26 = ( 2 + 13 ) + ( 11 )
27 = ( 3 ) + ( 5 + 19 )
28 = ( 2 + 7 ) + ( 19 )
30 = ( 2 + 3 + 5 ) + ( 7 + 13 )

A more interesting question is, for each natural number, how many distinct partitions into unique primes that include all prime factors there are. Note that zero can be partitioned into the empty set, so it has more solutions than 1.
I can give the min. value & it is one.
regards.

13. Originally Posted by rpenner
Ah, that was the piece I needed.

Shorter, stronger claim:
"All natural numbers other than 1, 4, 6, 8, 9, or 22 can be partitioned into unique primes such that all of their prime factors are present in the partition."
1 can't be partitioned into primes
Any prime can only be partitioned one way into primes including its factors, i.e., into itself
4, 6, 8, 9, 22 can't be partitioned into unique primes once their prime factors are accounted for.
Here is an (updated) list of how some small composite numbers could be partitioned:
10 = ( 2 + 5 ) + ( 3 )
12 = ( 2 + 3 ) + ( 7 )
14 = ( 2 + 7 ) + ( 5 )
15 = ( 3 + 5 ) + ( 7 )
16 = ( 2 ) + ( 3 + 11 )
18 = ( 2 + 3 ) + ( 13 )
20 = ( 2 + 5 ) + ( 13 )
21 = ( 3 + 7 ) + ( 11 )
24 = ( 2 + 3 ) + ( 19 )
25 = ( 5 ) + ( 3 + 17 )
26 = ( 2 + 13 ) + ( 11 )
27 = ( 3 ) + ( 5 + 19 )
28 = ( 2 + 7 ) + ( 19 )
30 = ( 2 + 3 + 5 ) + ( 7 + 13 )

A more interesting question is, for each natural number, how many distinct partitions into unique primes that include all prime factors there are. Note that zero can be partitioned into the empty set, so it has more solutions than 1.
Since 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.

14. Any counter example to the above statements?
regards.

15. Originally Posted by indianmath
Any counter example to the above statements?
regards.
indianmath, your last two statements have an "ad hoc" flavor to them; I get the impression that you made them after checking the first few dozen elements or so..?

I used to study number theory a bit and something that I didn't appreciate were conjectures that seemed to rely on statistical tendencies, as opposed to resting squarely upon deeper, hidden conceptual foundations.

Take Goldbach's Conjecture, for example: To find a counter-example we would have to produce a number n such that (x) and (n-x) are both composite for all x < n/2. The Prime Number Theorem states that there are roughly

$\frac{n}{ln(n)}$

primes less than n. The "blind odds" of either (x) or (n-x) being composite are then

$(1 - (\frac{n}{n*ln(n)})^2)$

and therefore the "blind odds" of finding a counterexample to Goldbach's Conjecture for all x < n/2 are roughly

$(1 - ({\frac{n}{n*ln(n)}})^2)^{\frac{n}{2}}$

At n = 100, this number is only 9% and at 10000 this number drops to $1.77*10^{-26}$! Therefore, Goldbach's Conjecture could very well be true solely due to statistical curiosities rather than being indicative of any fundamental law...

16. I agree that indianmath is over-confident that there are no counter-examples.
To do that, he would need to find a proof.
Nevertheless, no-one has yet shown that his idea is not new, or that it is wrong. So it is very interesting.

@Indianmath
I get the feeling for some reason that you are a young person still in education.
What stage of education are you at?

17. Originally Posted by Captain Kremmen
I agree that indianmath is over-confident that there are no counter-examples.
To do that, he would need to find a proof.
Nevertheless, no-one has yet shown that his idea is not new, or that it is wrong. So it is very interesting.

@Indianmath
I get the feeling for some reason that you are a young person still in education.
What stage of education are you at?
I'm not over-confident I'm saying it confidently that since all the above things are true therefore
" Any natural number except (4, 6 & certain prime numbers ) can be expressed as summations of at least Two distinct primes".
I think this statement proves Goldbach's conjecture also side by side.
regards.

18. Have you got a mathematical proof for that.
So far you have asked for exceptions.
As you know, that is not a proof.

You seem very confident that it will never fail at a higher number.
What gives you that confidence?
The next number it fails at could be very high.

Up to what number have you checked?
Or have you got a proof that it will never fail?

19. Originally Posted by Captain Kremmen
Have you got a mathematical proof for that.
So far you have asked for exceptions.
As you know, that is not a proof.

You seem very confident that it will never fail at a higher number.
What gives you that confidence?
The next number it fails at could be very high.

Up to what number have you checked?
Or have you got a proof that it will never fail?
It will never fail & I didn't asked for any counter examples of the post of my last statement made.
regards.

20. That's fine. Let's not argue.
I could be wrong saying that you need to provide a mathematical proof in order to say there are no further counterexamples.
Or perhaps you already have, and I didn't catch it.

I'll leave it to Rpenner.
He understands this stuff far better than I do.
If he doesn't think there is a problem, there probably isn't.

Personally, I would wonder why there are exceptions at 9 and 22, and then no further exceptions.
I would expect more. Not sure why.

Even if there were more exceptions, that would not make your idea less interesting.
It would provide a new series of numbers for people to ponder on.

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