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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.

I'd imagine that the proof that any number can equal any other number through dividing by 0 would've been like saying 2=3 because 2*0=3*0. As for the rest of what you said, I see your point.

When most people try to divide by 0, they either type it in to a calculator to find error, work it out on a sheet of paper to find repeating infinity (and I know writing infinity outside of calculus is against the rules, but it doesn't hurt to write it down just so you can see what will happen), or use calculus to get 'undefined'. Well, I believe that dividing by 0 is more of a philosophical process rather than a mathematical one. For instance, let's try using words to divide 10 by 2. To have one of 2 equal pieces to 10 is to have: 5 That seems to work out nicely. Let's try this with 0. To have one of 0 equal pieces to 10 is to have: Well, that's an invalid statement in that you can't have one of zero. I'm thinking that, when you try to divide by a nonzero number, it's like filling a glass with milk; but when you try to divide by 0, it's like trying to fill a glass that isn't there. You can spill the milk all over the counter, your floor, and even flood your house with milk all you want, but you'll never get any closer to filling that nonexistant glass. I believe that this is why, when we try to divide by zero on a sheet of paper, we usually end up with the largest number we can think of. Now, when most people set out to divide by zero, they normally look for a quantity. Now, imagining the empty glass as the essence of a quantity, I'm going to say that, when you divide by zero, your answer will be the absence of a quantity. This doesn't mean it's 0, because 0 is a quantity just like any other number. The absence of a quantity simply means you don't have a number, so your answer is literally nothing. However, let's look at this differently. One can always say that they have 0 of something. For instance: I have 0 ferraris parked in my driveway. I currently have 0 dollars in my back pocket. I have 0 solid gold keys that can unlock the doors to a solid gold house. Keeping this in mind, can we replace the absence of a quantity with 0? Well, I can't think of any instances where one would need to divide by zero except for one, negative factorials. 5!=5*4*3*2*1=120 4!=5!/5=24 3!=4!/4=6 2!=3!/3=2 1!=2!/2=1 0!=1!/1=1 -1!=0!/0=0 (Let's just assume for the sake of the test.) -2!=-1!/-1=0 -3!=-2!/-2=0 -4!=-3!/-3=0 -5!=-4!/-4=0 Now, when one uses factorials, they usually use it to see how many ways they can arrange n number of objects. For instance: I can arrange two objects 2 ways: GH & HG, which equals 2! I can arrange one object 1 way: G, which equals 1! I can arrange the absence of objects 1 way, which equals 1! Now, let's think about the negative numbers I've added to the pattern. If we assumed that you would get the absence of a quantity every time you divided by 0, and assumed that you could always replace that absence with 0, then all of the negative factorials would equal 0. I believe this is a valid answer. Consider holding negative five sticks of wood. You simply cannot do it, right? You can write negative signs on sticks as much as you want, but in reality, those are only positive sticks with negative signs written on them. Similarly if you're holding a handful of bills that says you owe money. You may think of it as negative currency, but they're just positive papers with numbers written on them. Since you cannot physically hold any number of negative objects, I believe it's reasonable to say that any negative factorial will equal 0. Since, because you cannot place negative objects on a table and shift them around, there will be no ways to arrange them. If anyone can think of any other situations where dividing by 0 is necessary, then please tell me. It would be a good way to prove/disprove my idea. Thanks.

Ah, I see now. If the most recent question I asked is answered with a yes, then that means it would be virtually immune to ciphertext only attacks, but not to attacks that compares known plaintext to the ciphertext (not unlike how the enigma flaw was exploited in nazi Germany's enigma code). Thanks for the information. I'll try and see if I can build something with it.

I think I now see why our thoughts on this are different. You're claiming that one cannot reach a solution that makes sense if they guessed the base value incorrectly, while I'm saying that it may be possible to reach an incorrect solution that can be read as if it where plaintext if one guessed the base value incorrectly. Now we just need to ask: "Which one of us is right?" Well, I believe that with a large enough base value, getting an incorrect answer that seems correct is possible. A good way to prove/disprove this would be to take the ciphertext I stated in my first post (1351478792050701588315749385525247210), take the key that's attached to it (both the scrambled letters and the base value) and see if we can turn the ciphertext into a different readable message by only changing the key ( changing the message into something that isn't Hello_World!).

Yes, but this would only work if there was only one readable message you can get from using the method you've stated. I'm going to turn to the one time pad for an example. Let's say that Eve intercepts the one-time pad cyphertext message Alice sent to Bob and attempted a cryptanalysis on it. Assuming Eve has an infinite amount of computing power, she can figure out the key and get the plain text instantly. However, she notices that one key and the plaintext HELLO is just as likely as a different key and the plaintext LATER. I'm trying to replicate this feature and put it in to a more usable cryptosystem. Now, there's another question I need an answer to. Does having more pseudorandom numbers make determining the pattern between the numbers easier to find? I've read from somewhere that a major problem with the one-time pad is that it relies on the randomness of the numbers in the key, and that making a truly random set of numbers is a non-trivial problem. Would having less pseudorandom numbers make finding the pattern between the numbers harder to find? I'll be sure to give this a read. However, I think you should read the reply I sent to Cap'n Refsmmat. It's basically the same thing I planned on telling you, but stopped since repeating myself would unnecessarily take up more space.

A random number that, when used as a base value, can hold all of the types of symbols you plan to put in to the text. For instance, if you where to use the alphabet and a space, then you would need to use a minimum base value of 27, or 0 to 26. You will also need to assign the symbols numbers on the base value at random. The key you want to share to your recipient and keep a secret from everyone else is the base value and how the symbols (letters, numbers, and punctuation), are spread across the base value However, since using the absolute minimum base value would be the same as scrambling the letters with extra calculations involved, you'll want to use a base value that is much larger and completely random. So basically what you're saying is that, by getting in to the recipient's system, you can take the key from him/her. Assuming that I'm following correctly, I'm going to ask a question. Would creating a public-key version of this fix the problem?

I believe you're forgetting the fact that the base value is also kept a secret and is a part of the key. I want you to pretend that you don't know the base value is 1337, and then try again.

Hello_World! KEY~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ The base is 1337 (0 to 1336). H=55 e=509 l=1205 o=798 _=319 W=562 r=900 d=29 !=1329 KEY~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 55*1337^11+ 509*1337^10+ 1205*1337^9+ 1205*1337^8+ 798*1337^7+ 319*1337^6+ 562*1337^5+ 798*1337^4+ 900*1337^3+ 1205*1337^2+ 29*1337^1+ 1329*1337^0 1351478792050701588315749385525247210 1351478792050701588315749385525247210/1337=1010829313426104404125467004880513 R1329 ! 1010829313426104404125467004880513/1337=756042867184820047962204192132 R29 d 756042867184820047962204192132/1337=565477088395527335798208071 R1205 l 565477088395527335798208071/1337=422944718321262031262683 R900 r 422944718321262031262683/1337=316338607570128669605 R798 o 316338607570128669605/1337=236603296611913739 R562 W 236603296611913739/1337=176965816463660 R319 _ 176965816463660/1337=132360371326 R798 o 132360371326/1337=98998033 R1205 l 98998033/1337=74044 R1205 l 74044/1337=55 R509 e 55 H Hello_World! I'm sure you can see what I'm trying to do. I'm basically taking a bunch of letters, placing them randomly on to a random base value, treating the plaintext as if it where a number on that base value, and translating it into decimal form. I believe that this is unbeakable in a similar sense to the One-time pad, in that, with unlimited computing power, one can generate the key instantly, but would find that the key is lost inside an infinite number of other key possibilites. Is what I'm saying true? Another thing I would like to have clarified. I understand that one does not simply generate random numbers on a computer, and that the numbers would only be pseudorandom; but would it be harder to find patterns in pseudorandom numbers if there where less numbers to work with? Thanks.