The curiousity that I am about to present I discovered after trying to solve an interesting puzzle that proceeds as follows...

There are 2 mathematicians S and P. S knows the sum of two numbers and P knows the product of those same two numbers. Assume that the numbers are integers greater than 1 and that the mathematicians know this. For simplicity, let the first number be less than or equal to the second. Furthermore assume that each mathematician only speaks truly. The conversation goes as follows:
S: I know that you don't know what the two numbers are.
P: Now I do know what the two numbers are.
S: Now I know what the two numbers are.
What are the two numbers?

(I am told that this problem was originally proposed by Dr. David Sprows of Villanova University)

The first piece of information that S gives us is that the sum of the numbers is not something that can be expressed as the sum of two primes. For instance, S could not make this statement if he knew the sum to be 19 because for all he knows P knows the product to be 34 in which case P would know what the two numbers are, namely 2 and 17.
So S's statement has content and furthermore seems to be enough to give P the answer. Must the sum be an odd number. It seems so because even numbers have many ways to be the sum of two primes. A number like 36 can be expressed as 5+31 or 7+29 or 13+23 or 17+19 and it seems that the higher the even number the more pair of primes that add up to it.

But odd numbers either have one way or no ways to express them as the sum of two primes. In fact when the odd number can be expressed as the sum of two primes it is because the lowest number is a 2. So I reasoned that the content of S's first statement was that the sum was not a prime plus 2 but it was an odd number. And from there I went on to solve the problem, which is another paper.

But then later I worried about my assumption that every even number could be expressed at least one way as the sum of two primes. (I later discovered through responses to a post like this that such an assumption is known as Goldbach's conjecture.) My reasoning was inductive at best and seem to leave open the possibility that maybe there was some large even number that could not be expressed as such a sum. (You may ask how intense such a worry could truly be). So I decided to write a program that would tell me how many ways there were to express the even numbers between 2-100000 as the sum of two primes (not including 1 as a prime). So ,for instance when the program came to the number 36 it discovered that there were four ways to sum up 2 primes to 36, the four I mentioned earlier. 5+31, 7+29, 13+23, 17+19.
So then I plotted of the number of ways to sum as a function of the even numbers and got the following graph. Hoping that it would give me some insight into how I could assure myself that there was no even number unexpressible as the sum of two primes.

The graph did what I expected, in one sense. It seemed to confirm that the greater the even number the greater the number of summing prime factors, on the average, that is. And while it gave me no insight into the proof that it would always be such, it presented me with another strange and interesting problem. Why wasn't this graph more random looking? Why are there streaks of concentrated points? Some streaks twice as high as others?


Looking at the actual data gave me a clue. I then wrote another program, one that colored the even numbers depending on their factors. For instance if a number was divisible by 3 but not 5 or 7 then its was colored yellow.

This was the color code
Yellow: Divisible by 3, but not 5 or 7
Green: Divisible by 5, but not 3 or 7
Cyan (light blue): Divisible by 7, but not 5 or 3
Blue: Divisible by 3 and 5, but not 7
Magenta: (weird red): Divisible by 3 and 7, but not 5
Red: Divisible by 5 and 7, but not 3
Black: Divisible by 3, 5 and 7
Black again: Not divisible by 3, 5 or 7

I received the following graph

The high very thin scatter of black points represents the even numbers that are also divisible by 3, 5 and 7 and the low black streak represents the not divisible by 3,5, or 7 numbers. The higher red is the magenta divisible by 3 and 7, the lower red is the ones divisble by 5 and 7.


It seemed clear then that the more different divisors a number has, the more likely it is that it have many pairs of primes that sum to them, and furthermore, numbers with lower divisors have more ways than numbers with higher divisors. (all things being equal and expressed as an average)

I suspect that this has all been noticed before, but if it has not, then let me submit it and you may do with it what you might. I don't really understand why a number with a lot of prime factors should also be a number that can be expressed many ways as the sum of primes but that seems to be the case. If you have some ideas on this, please let me know. I would be very interested.

By the way, the answer to the original puzzle is 4 and 13.***

I don't know how old your "Curious primes" webpage is, or if you're
still interested in the question, but (while I'm not an expert in
number theory) I think I can give an explanation for the phenomenon
you observed -- though it's a heuristic argument rather than a
complete proof, since a complete proof would probably either resolve
the Goldbach conjecture, or come a lot closer to it than I'm able to.

The heuristic is that in general, the distribution of primes "looks"
fairly uniform with respect to most obvious measurements. Sometimes
this can be formalized, e.g. in variants of the prime number theorem
that show that for a fixed prime p, if we take all primes up to some
(sufficiently large) integer n and associate them with their values
modulo p, each value modulo p will correspond to about the same number
of primes. But similar things seem to hold even when we can't prove
it, e.g. the density of twin primes (primes at distance 2 from each
other) seems to be about the square of the density of primes (i.e. the
probability of n and n+2 both being prime is approximately the same as
if we just considered the individual probabilities as though they were
independent) -- though this is unproven, since it is still an open
question whether there are even an infinite number of twin primes.

The application to your observation is this: for an even number n, the
number of ways n can be written as the sum of two primes can be found
by starting at n/2 and successively checking whether (n/2 + i) and
(n/2 - i) are both prime, for all i up to n/2. By the uniform
heuristic above, all else being equal, we would expect the probability
of (n/2 + i) and (n/2 - i) both being prime to be roughly the product
of the probabilities that the individual numbers are prime.

Now, suppose n is divisible by (say) 3. Then n/2 is also divisible by
3. We then know that (n/2 +- i) cannot be prime when i is also
divisible by 3, so the individual "probabilities" in that case are 0,
while for i not divisible by 3 the individual probabilities will be
more or less uniform (by the variant of the prime number theorem I
mentioned above, though the application here has a noticeable error
term because of the way I'm using it).

On the other hand, suppose n is *not* divisible by 3 -- let's say for
concreteness that n/2 is 1 modulo 3. Now, if i is also 1 modulo 3,
then (n/2 - i) is divisible by 3 -- the "probability" of its being a
prime is 0, hence so is the probability that both (n/2 - i) *and* (n/2
+ i) are prime. On the other hand, if i is 2 modulo 3, then (n/2 + i)
is divisible by 3, and the probabilities are again 0. This leaves only
the 1 in 3 case that i is divisible by 3 where we can even hope to
find primes at both (n/2 - i) and (n/2 + i).

This means, roughly, that if n is divisible by 3, we must immediately
eliminate 1/3 of the possible sums from consideration, whereas if it
is not, we must immediately eliminate 2/3. lowering the expected
number of valid sums.

Similarly, if n is divisible by 5, we immediately eliminate 1/5 of all
possible sums, whereas if it is not, we eliminate 2/5, and so on for
all primes up to n/2.

This argument depends entirely on the informal claim that all of these
"probabilities" of a number being prime will be fairly uniform,
independent of all the different individual cases. The upshot is that
the ideal case is where n/2 is a product of very small primes, so we
"throw out" as few primes as possible.

Anyway, I don't know how helpful this is -- since even the experts
seem to think the Goldbach conjecture is pretty uncrackable with the
current state of knowledge, I'm not sure what the odds are of
formalizing any of the preceding. But this is at least my intuition
for why you got the results you did.

David

To the mailbox at billtomlinson@mac.com

Back to my homepage

***

After having this page up for 2 years, Gordon Berg e-mailed me that there are actually many more solutions compatible with the 2 Mathematician puzzle. I have gotten responses like that before that I was easily able to refute so I had initially assumed that he had overlooked something. But after a few communications I started taking his claim more seriously. Seriously enough to write a new program that found new solutions that he also had.

He told me that there are five others between 17 and 177 - 65 (4+61), 89 (16+73), 127 (16+111) , 137
(64+73) and 163 (32+131) and that there is no indication that the density of solutions decreases. Because I wrote a program independent of his and got the same numbers, this constitutes assurance (at least for me) that the new result is correct. See what you think. Meanwhile there is a way to modify the puzzle to make the solution unique and maintain a generality.

For instance,

There are 3 mathematicians S, P and U. S knows the sum of two numbers and P knows the product of those same two numbers and U knows a number x that he also knows is greater than the product (knows an upper limit on the product), although he doesn't know the product. Assume that the numbers are integers greater than 1 and that the mathematicians know this. For simplicity, let the first number be less than or equal to the second. Furthermore assume that each mathematician only speaks truly. The conversation goes as follows: