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Chen’s Theorem and YAPFO September 15, 2016

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Chen’s Theorem

This posting is about the Chen’s Theorem project.

Let’s start out by simply stating Chen’s Theorem.

 

In number theory, Chen’s theorem states that every sufficiently large even number can be written as the sum of either two primes, or a prime and a semiprime (the product of two primes). For example:

20 = 17 + 3      (2 primes)

20 = 5 + (3 x 5)       (a prime and a semiprime)

 

The part of the theorem about the sum of 2 primes we’ll ignore here because it’s the same as Goldbach’s Conjecture which was covered in an earlier posting

(see Gerry’s Goldbach Theorem)

 

The part of Chen’s Theorem we will cover here is the part about:

or a prime and a semiprime

 

The initial process of exploration used the usual “method.” I.e. “manually” play around with a few randomly chosen example numbers on a white board (EVEN numbers in this case) and try to spot patterns. If a meaningful/interesting pattern surfaces and it looks to be worth pursuing then the next step is to decide whether to write a computer program to test the perceived pattern over millions (or billions) of numbers to see if it still holds water. Writing the program(s) is never difficult but it’s always time-consuming; and I don’t want it to interfere with my watching of TMZ or AGT!!

Even if a pattern seems to hold true over billions of numbers, and even if some of those numbers are hundreds of digits long, it still doesn’t prove anything! It’s still just conjecture. We learned that from experience and it was touched on in a prior posting:

See the SumOfPowers article

Look for “Lander” or “Parkin” in the above posting/article.

 

Anyway… Even if the perceived pattern holds true across billions of tests, it’s still just conjecture. The next step is to actually Prove the truth (or falsity) of  our conjecture.

A perceived benefit of writing code to run thru billions of test cases is:

If our conjecture is false, we can greatly increase our chances of quickly realizing it by way of counter-example. We could save a lot of effort by not trying to prove something that can not be proved! Again, Look for “Lander” or “Parkin” in the posting/article SumOfPowers.

Of course, it may be that proving the conjecture could turn out to be easier than we think. If we attack the proof early on we could, perhaps, prove (or disprove) our conjecture with relative ease and thereby save ourselves the programming effort.

 

All that said…

The whole “Proof process” seems to have been getting easier lately so I’ve taken to trying to Prove the conjectures (pattern perception) in advance of writing any code to test billions of specific examples. And keep in mind that I would eventually have to go thru the whole Proof thing anyway! Why not at least give it a shot early on!? It might turn out to be relatively simple and I could save myself a lot of time.

So… for this project the following was the process…

 

Randomly choose an EVEN number that would be easy to work with via calculator (or in my head) and apply the following part of Chen’s Theorem:

the sum of a prime and a semiprime For example (and color coded):

The prime factors of 20 are 2, 5

20 = 5 + (3 x 5)            (sum of a prime and a semiprime)

20 = 11 + (3 x 3)

or

The prime factors of 68 are 2, 17

68 = 3 + (13 x 5)

68 = 11 + (3 x 19)

68 = 13+ (5 x 11)

68 = 17 + (3 x 17)

or

The prime factors of 12 are 2, 3

12 = 3 + (3 x 3)

12 = 2 + (5 x 2)

ETC.

 

The pattern noticed (i.e. the conjecture), is that, whenever one of the prime factors of the EVEN number on the left side of the equals sign shows up on the right side of the equals sign, then that prime factor will be in BOTH terms on the right of the equals sign. That is, that prime factor will constitute BOTH the prime, and, one or both of the primes that makes up the semiprime as shown in the examples above and color-coded in red.

Clearly it would be easy to write a program to spin thru bizillions of EVEN numbers to test the above conjecture. And if we were to encounter even a single counterexample it would be enough to PROVE the conjecture is false. But even if all of the examples tested conformed to the conjecture it would not be enough to actually PROVE it.   It would lend credence to the conjecture but it is not PROOF.

 

For this project I decided to try proving the conjecture in advance of any programming.    As it turned out, the proof came quickly and is very simple. Here it is; I call it:

 

Gerry’s Chen Based

YAPFO Theorem

Proof

YAPFO = Yet Another Prime Factor Oddity

 

 

First we restate Chen’s Theorem upon which this YAPFO Theorem is based:

In number theory, Chen’s theorem states that every sufficiently large even number can be written as the sum of either two primes, or a prime and a semiprime (the product of two primes). For example:

20 = 17 + 3 (2 primes)

20 = 5 + (3 x 5) (a prime and a semiprime)

The part of the theorem about the sum of 2 primes we’ll ignore here because it’s the same as Goldbach’s Conjecture which (again) is covered in an earlier posting (see Gerry’s Goldbach Theorem);

 

The part of Chen’s Theorem of interest here is the part about:

“or a prime and a semiprime “

Stated symbolically:

E = P + (Ps1 x Ps2)

where P is some prime, and, Ps1 and Ps2 are primes that make up the semiprime (Ps1 x Ps2). And E is, of course, some EVEN integer greater than 4.

The proof proceeds as follows:

E = Pf1 x Pf2 x Pf3… x Pfn

where Pf1 .. Pfn   are the prime factors of E.

Which leads to…

Pf1 x Pf2 x Pf3… x Pfn   =    +   (  Ps1 x Ps2  )

 

If we divide both sides by any one of E’s prime factors Pf1 .. Pfn   we get (using Pf1 for example)

 

                              P            ( Ps1 x Ps2 )

  Pf2 … x Pfn   =         ——    +     ———————–

                             Pf1                     Pf1

 

Note that the left side of the equal sign is an integer. But also note that:

If   P = Pf1  then for the right side of the equation to be an integer,  Pf1  must also equal Ps1 (or Ps2… it doesn’t matter which).

And likewise, if Pf1 equals Ps1 (or Ps2),   then in order for the right side of the equation to be an integer, Pf1 must also equal P.

In other words,    Pf1  can not evenly divide ( Ps1 x Ps2 )   unless Pf1 equals Ps1, or  Ps2, or both.  

 

So… for       E = P + (Ps1 x Ps2)

then if one of the prime factors of E equals any of the terms on the right side of the equals sign,  then it must equal at least 2 of the 3 terms,

one of which must be  (Pf1 must equal P).

 

By the way, here are some examples where one of the prime factors of E is equal to all 3 terms on the right side of the equal sign (they are easy to construct):

12  =  3 + (3 x 3)

6   =   2 + (2 x 2)

30 = 5 + (5 x 5)

 

 

 

*** end of proof ****** end of proof ****** end of proof ***

 

Ta Da!

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Now for an interesting (hopefully) note about the history of this short project…

It’s been awhile since working on a new project. I’ve been on the lookout for one (as always) but only a couple of days ago did I find Chen’s Theorem. Interestingly, while looking for a project subject, I thought that what might help is a central source for Number Theory formulas and theorems. I also thought “What took me so long to figure that out!?”

So I searched for that and voila. I immediately found the following:

https://en.wikipedia.org/wiki/Category:Theorems_in_number_theory

A valuable  find.

 

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