Cosmology

[desiresjab]

I trust algebraic proofs as much as I trust the results of a computer program. There could be something wrong in either that has little to do with the original problem. Being convinced is the intuitive part of the proof.

One way that I think about factorials is to view them as the number of ways to order n objects. In the first position one could have n choices. After that choice there are n - 1 choices for the next position and so on all the way down to the last piece and there is only one way to order it. Multiply all those together and you get n! For the multinomial formula coefficient the numerator is the way to order n objects. The denominator is the way to order subsets of those objects separately.

I was looking at Carmichael's "Theory of Numbers" and he approaches the problem by considering the highest power of a prime p in n! which I am not familiar with but sounds interesting. Then if the highest power of a prime in the denominator is less than or equal to the highest power of the prime in the numerator the fraction would be an integer. This might be more useful than the multinomial coefficient approach since it could answer more questions.

I agree that it is good to find multiple ways to solve something. The easier, the better.

Could not agree more. I have looked at the link now and what we have here is the same recipe taken out of context. Really, every time one is working with factorial formulae it would be wise to check for this connection.

Carmichael may be using Polignac's formula. That is what it does, I believe.

Factorials are beasts. It is easy enough to see what they are and what they designate for counting (very good description of their role in the multinomial, by the way), but when you want to compare them to something like a power it becomes sticky. One reason I appreciate Wilson's theorem so much is that it relates factorials to powers. That is cool.

It is time to move on. My own urge is toward Borcard or Goldbach. I have learned a lot from Brocard's problem. The Goldbach conjecture, however, is difficult to make even a quarter inch of progress on. I suppose a place to start would be with Ramanujan's amazing formula for summing additive partitions. No one believed such a formula was even possible.

When looking at problems which minds like his have already considered, one can only hope they were in a great hurry that day.

[desiresjab]

If all the mysteries of a simple number line were solved, the human race might become more advanced than the alien civilizations we envision in our fiction. Complete power over a number line would be god-like. The solving of some problems would contribute mightily to acheiving this power. The Reimann conjecture is high on the list. I think of Goldbach's conjecture as very important.

The twin prime conjecture is one that fascinates people. We know something about twin primes from studying QR on here for so long. Since twin primes are a mixed couple in terms of 4n+1 and 4n+3, their combined Legendre symbols will be (1)(-1)=-1.

Working on unsolved problems there is generally little or nothing to report. Many a doctoral dissertation in math has explored some tiny area of these problems. Any progress is counted a success.

[YesNo]

I didn't see a reference to Polignac, but it looks like the same argument after checking Wikipedia.

We should probably just assume the Riemann hypothesis is true. Then derive some consequences from it and check if they are true or not. Inadvertently someone might discover that it is false by finding a consequence of it that is false.

I am still working on the Sierpinski problem. Or rather, I think about it off and on. There isn't a lot of work getting done.

[desiresjab]

The Reimann conjecture may be overrated in terms of what the impact of its solution would be. I do not know for sure. Researchers can already assume it is true and take it from there. That is done with a lot of propositions. I believe it is usually called the "weak form," if a proof in the chain is missing. Can anyone say what else would be immediately true if the Reimann hypothesis were true? Perhaps it is a hugely significant problem on its own.

It seems to me that a proof of Goldbach's conjecture would almost necessarily lead to a fuller understanding of the additive nature of primes. Is there even such a theory to be had? What the mathematical world needs right now is Ramanujan. If Goldbach is solvable, I think Ramanujan had the best chance. Had he lived beyond his twenties it might be solved now.

There are two biographies of mathematicians I can recommend. The Man Who Loved Only Numbers and The man Who Loved Only Infinity, about respectively Paul Erdos and Ramanujan. They are both fantastric reads in my opinion.

I turned back to Brocard's problem and immediately made progress on an area of it I was not equipped to several years ago. I am going to hang with it again for a while and see if anything else pops out for me. Originally, I said all I wanted to do was see if I could get to the same vista where Ramanujan had stood before he gave up on this problem. Perhaps I have done that, for the present stage seems impassable. But of course no one knows where Ramanujan stood, so I must trudge onward.

[YesNo]

According to Wikipedia, Brocard's problem (https://en.wikipedia.org/wiki/Brocard%27s_problem) is related to the "abc conjecture": https://en.wikipedia.org/wiki/Abc_conjecture. However, I don't see the connection at the moment.

This article searching for solutions was also cited: http://www.math.uiuc.edu/~berndt/articles/galway.pdf

There is also some discussion on stack exchange: https://math.stackexchange.com/searc...rd%27s+problem

[desiresjab]

I can see where it might relate. Almost everything in number theory seems to relate anyway, and it is still Diophantine equations. They also use Legendre symbols from QR in the research. Since so much relates, that is what the smart guys have been doing interrupted for a couple of hundred years, which is exactly the reason and the only reason we know the importance of particular unsolved problems in immediately solving other unsloved problems. It is amazing how the brilliant boys and girls keep knocking chips off these problems until someone gets a finished sculpture. They do it with brilliance in various areas of high math. The Japanese who claims a proof of the abc conjecture invented entirely new methods, from the reports, which went far outside number theory, perhaps any theory. I still have to look at it to see if I can make monkey of anything he says.

In the meantime, I trudge along. I am within sight of a new perch from which to see the problem.

[desiresjab]

Most of the progress I thought I made on Brocard was illusory, further thought showed, though there was a bit of an increase in understanding. Still, I am essentially where I was last time I left on working on it after all. I feel a real shortage of tools. The problem with known tools is that everyone who is greater than me has tried them.

[YesNo]

There is a proof of the abc conjecture that people are trying to verify although the proof is very long: http://phys.org/news/2016-08-abc-proof.html

If that proof convinces others then understanding it may be more important than Brocard's problem. However, I am still trying to understand why the abc conjecture is relevant to Brocard's problem.

[desiresjab]

There is a proof of the abc conjecture that people are trying to verify although the proof is very long: http://phys.org/news/2016-08-abc-proof.html

If that proof convinces others then understanding it may be more important than Brocard's problem. However, I am still trying to understand why the abc conjecture is relevant to Brocard's problem.

I am aware of that proof. It uses properties of number classes that are outside conventional mathematics. Elliptic curves are still generating solutions to unsolved problems. These guys are using group theory, abstract algebra, moduli and Elliptic curves and topology. His proof uses almost no calculus, so is elementary, which means anything but simple, as I have been trying to convince people for a long time.

Anyone who is not an expert in the above fields may as well forget actually understanding his proof. I know that leaves me out. Of course I will poke around with it anyway.

[desiresjab]

Along with all sorts of other strange mathematical objects and operations, such as rings, ideals, kernels and cosets, he uses something called theatres. This involves treating certain groups as if they were abstract topological fields without the labeling. Mochizuki provides this new labeling. There is a distortion during the operations of multiplication and addition in the ring that he measures and accounts for outside the group.

It does not pop out at me why the abc is linked to Brocard. But everytime I read about this stuff in depth I learn something new. For instance, I finally understand clearly what an algebraic integer is. Ideal rings are much clearer now too. Immediately I see a connection of theirs to Brocard. Sorry I have to be so cryptic. You see, I am slightly luny and still intend to solve this thing before the abc conjecture does it sweepingly!

It is truly amazing to me how far number investigations have been taken. Now Mochizuki has added some new beasts to the zoo.

[YesNo]

Based on the likelihood that the abc conjecture is true, then the number of solutions to Brocard's problem is finite. The only question remaining is to either show that all the solutions have already been found or to find another one which should have more than 20 digits. Another contribution would be to find an algorithm faster than computing quadratic residues to check that a solution does not exist.

[desiresjab]

Based on the likelihood that the abc conjecture is true, then the number of solutions to Brocard's problem is finite. The only question remaining is to either show that all the solutions have already been found or to find another one which should have more than 20 digits. Another contribution would be to find an algorithm faster than computing quadratic residues to check that a solution does not exist.

I have an approach to Brocard I have not seen per se. An instinct tells me it is solveable and that I am on a good course. I chose this problem long ago because its shape was pleasing to me. I like factorials. Right now I am at an impasse, looking for a way around. One is always at an impasse on unsolved problems, then suddenly a little progress is made. Often these seeming advances are illusory, the result of a mistaken notion one realizes later. So progress really is slow, but that makes any advance exciting.

[YesNo]

It is always good to come up with a simpler solution than the ones known.

Regarding the abc conjecture, I can see how the form relates to Brocard's problem with is n! + 1 = m². Here a = n!, b = 1 and c = m². The product of all distinct primes in the product abc is close to c, that is the product of the distinct primes in this product (n!)(1)(m²) is about m². I can see that m would be larger than n, but why wouldn't it have some primes in common with n!?

[desiresjab]

It is always good to come up with a simpler solution than the ones known.

Regarding the abc conjecture, I can see how the form relates to Brocard's problem with is n! + 1 = m². Here a = n!, b = 1 and c = m². The product of all distinct primes in the product abc is close to c, that is the product of the distinct primes in this product (n!)(1)(m²) is about m². I can see that m would be larger than n, but why wouldn't it have some primes in common with n!?

Because m² being one greater than n! can share no factors with it. Euclid used the same idea in his proof of the infinitude of primes.

You set Brocard into the form correctly. I still cannot see why it would have an impact on Brocard's problem, because the task there is to show if the difference between a square and a factorial can ever be exactly 1, other than the three known cases. Researchers are trying to bound the function from above.

There is a relationship between factorials and triangular numbers I find fascinating.

(2n)!=2ⁿ _k=1∏_n T_2k-1.

Stunning!! What this really says is: factor out n powers of 2 from (2n)!, then multiply it by all the oddly labled triangular numbers which have been multiplied together all the way up to one less than twice the value of n, and you will be back at 2n factorial.

[YesNo]

Right. I can see now that m² must be relatively prime to n! because any factor of n! that divides m² must also divide 1.

The relationship with triangular numbers is interesting.