 # Is There a Possible Connection Between Brown Numbers and Mersmenne Primes?

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Okay, so I was reading about Brocard's Problem, and how there are only three known whole number solutions to it. To anyone who doesn't know what I'm talking about, this is the problem:

n!+1=m^2

The only known pairs of numbers that will satisfy this question while staying whole and valid are (4,5), (5,11), and (7,71). Two men, Berndt and Galway, also performed calculations up to one-billion for n and showed that there where no further answers to satisfy this problem within that range. Paul Erdős has even conjectured that no other solutions exist!

Naturally, when somebody reads something like this, they want to look for more solutions. I figured that, since calculation up to one-billion where already performed, I would have next to no luck in finding more brown numbers (whole number solutions that satisfy the equation) by randomly picking points on a graph. So I figured that I should start looking for patterns. Even if the quantity of numbers that I have to work with is quite humble, it's better than complete guessing, right?

Let's get started. First off, most people usually focus around n, so for creativity's sake, we're going to focus on m. We can tell that 5, 11, and 71 are all prime numbers, but that's not enough to go on in my opinion. I decided to perform random actions involving these primes. Adding, subtracting, ect. Eventually, albeit hardly-any-better than our last observation, I found something else.

2^2+1=5

2^3+3=11

2^6+7=71

A lot of people would know that these two numbers are the first two mersmenne primes (2^2-1 & 2^3-1). I figured that, by combining this with our last observation that all of the known numbers for m where prime, we may be able to find a pattern that would allow us to find new brown numbers. My hypothesis is that the pattern would is 2^x+q=p, where p is a prime, q is a mersmenne prime, and x... Hmm, I can't really see a pattern in x. If you see one, please feel free to tell me. Other than that, we'll just push it off to the side for now.

So, if my hypothesis is correct, then the next number for m would be 2^x+31. I made a program for this on my mediocre desktop, but it ran out of memory before any solutions could be made where n!+1=(2^x+q)^2. If anyone here would like to have a go at it, be my guest.

On a side note, it's just as possible that this isn't following a pattern of mersmenne primes, but is instead simply following a pattern of 2^n-1.

If we where to speculate further, there is another question we could ask if my hypothesis turns out to be true. Overholt had stated that, if abc conjecture is true, then that would mean that there's finitely many brown numbers. If we where to a, prove that there where finitely many brown numbers; and b, prove a significant connection between brown numbers and mersmenne primes, then could we also prove that there are finitely many mersmenne primes? Of course, the foundation for everything I'm saying is very mushy (and nobody's above criticism), so please feel free to tell me what you think. ##### Share on other sites

Okay, so I was reading about Brocard's Problem, and how there are only three known whole number solutions to it. To anyone who doesn't know what I'm talking about, this is the problem:

n!+1=m^2

The only known pairs of numbers that will satisfy this question while staying whole and valid are (4,5), (5,11), and (7,71). Two men, Berndt and Galway, also performed calculations up to one-billion for n and showed that there where no further answers to satisfy this problem within that range. Paul Erdős has even conjectured that no other solutions exist!...

Let's get started. First off, most people usually focus around n, so for creativity's sake, we're going to focus on m. ...

Of course, the foundation for everything I'm saying is very mushy (and nobody's above criticism), so please feel free to tell me what you think. I think the biggest problem is in the computability of n! for large values of n. This is likely the wall that Berndt & Galway hit, because it's easy to check for a solution in m by taking its square root when you have the value of n!. Even if you start with m you would still have to compute the necessary n! to check for a solution. Offhand I'm not aware of any shortcuts to factorials. Good luck!

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