Strange Posted May 15, 2017 Share Posted May 15, 2017 On 5/14/2017 at 9:31 PM, Trurl said: As you said before a drawing will show everything. I don’t have any programs to draw triangles other than AutoCad 14. Microsoft Word? OpenOffice? LibreOffice? Online tools such as Vectr? Quote So, if you can view my drawing. It may be awhile before I have a drawing in universal format. You can't export as PDF, GIF, JPG, PS, PNG, TIFF ... ? Screen grab? https://forums.autodesk.com/t5/autocad-2000-2000i-2002-archive/exporting-to-a-gif-or-bmp/td-p/192390 http://www.zamzar.com/convert/dwg-to-gif/ Link to comment Share on other sites More sharing options...

imatfaal Posted May 15, 2017 Share Posted May 15, 2017 On 5/14/2017 at 9:31 PM, Trurl said: Obviously, you didn’t look at the drawing I posted. This is a science and engineering forum. There must be some viewer who has access to Autocad. ... I did look at the drawing. I think the triangles are not necessarily similar (ie there is a circumstance in which they are similar - but most of the time they are not). Drawings are very misleading if you use them to measure lengths and angles (as you are doing). You should use diagrams to get your thoughts straight but use geometrical rules to determine things like congruency. I can draw an equilateral triangle - with enough precision to be confident in it to the full extent of a rough diagram - in about twenty seconds. Take a picture with a smartphone. Upload it. On 5/14/2017 at 9:31 PM, Trurl said: ...I know you think, I’m stupid claiming to work with Prime numbers. And you don’t think I have a math background. I enjoy you reading my problem and telling me when something just doesn’t work. However, you should have viewed the drawing before dismissing my comments. If I am wrong your judgment is correct. However, there is no wrong. There are wrong techniques, but failures just mean that I try other approaches. In this problem, I am not asking you to find Prime numbers, I am asking the group to find techniques to find triangles with limited given. If you want to reverse a one-way function, you will have to use new ideas, because the old ones don’t work either... You are making more assumptions than I did - there is an autocad viewer in the windows 10 appstore. I viewed your diagram And there is a wrong. For example; one of yours - if a+b<c then a,b, and c CANNOT form a triangle and it is wrong to posit the triangle ABC. We know a huge amount about triangles, constructive and deductive geometry, and, to move on in complexity, trigonometry. The main problem is that you don't seem to have a good grounding in this. On 5/14/2017 at 9:31 PM, Trurl said: ...I understand why you think my problem is crap. But don’t think of it as supposed to solve Prime numbers. Think of it as a geometry problem where limited information is known about the angle. Yes, I could be wrong, but I believe I am right-on about the similar angles. The question is does it help me solve the unknown values of the triangle I need... But as a geometry puzzle it lacks precision - everyone reading it should be able to reproduce the construction easily but this is not possible because you chance the rules when some part is proven to be wrong Link to comment Share on other sites More sharing options...

Trurl Posted May 19, 2017 Author Share Posted May 19, 2017 You sound like you have some good math wisdom. I value your comments. I will have to evaluate the use of this geometric diagram. I didn't know there was a AutoCad viewer in Windows 10 store. I didn't know people actually used the store. I'll have to check it out. Anyways I was wrong. You did give a fair view of my problem. I misunderstood you comments about viewing the drawing. I have claimed in many posts ago on a different thread that I could solve SSA (side, side, angle). I did not want to bring that back up. But later I watch a math video on "The Great Courses" that the mathematician said he could solve SSA with some obtuse triangles. That is if I understood him right. This problem is not based on that. But again it is trying to find triangles knowing only 2 sides, with certain conditions known by other triangles. The following is not a religious statement. But in Christianity an non-believer would say if God is so strong can he make a boulder that he can't lift? If he can't he isn't all powerful. But you could extend it to this problem. If God is so powerful can he make a one-way function even he can't reverse? So do one-way functions exist? I believe as humans we are limited and one-way functions do exist. Knowing only N and finding 2 unknowns, knowing only a slight process in which N was encoded is a difficult if not impossible task. I have been working on this problem for a long time. I took a break to study Amateur Radio. But I would like to share the patterns in the multiplication I have found and based my equations. They are simplistic but I have to go through and list them all. Maybe someone will see something I missed. Anyways thanks for the comments. I don't always like being wrong, but I think it is more about being realistic with this problem. I don't know. Would you be interested in seeing a 5 step pattern in multiplication? Link to comment Share on other sites More sharing options...

imatfaal Posted May 20, 2017 Share Posted May 20, 2017 On 5/19/2017 at 7:22 PM, Trurl said: ...Anyways I was wrong. You did give a fair view of my problem. I misunderstood you comments about viewing the drawing. I have claimed in many posts ago on a different thread that I could solve SSA (side, side, angle). I did not want to bring that back up. But later I watch a math video on "The Great Courses" that the mathematician said he could solve SSA with some obtuse triangles. That is if I understood him right. This problem is not based on that. But again it is trying to find triangles knowing only 2 sides, with certain conditions known by other triangles. Yes - I understand that; you need to get your head around what can be done and what cannot. There are numerous texts on this sort of geometrical conundrum. Which angles are equal, supplementary, complementary and which lines are parallel. Most of this can be done with a careful diagram done with ruler and compass. You can get to trigonometry and cosine and sine rules - but then you also have to consider that triangles with whole number sides (primes remember) are a requirement. Have you actually tried constructing triangles with known primes - you obviously haven't because your early assertions to the numbers required were impossible. I presume you know how to construct a rough triangle with a straight edge, ruler, and compasses On 5/19/2017 at 7:22 PM, Trurl said: The following is not a religious statement. But in Christianity an non-believer would say if God is so strong can he make a boulder that he can't lift? If he can't he isn't all powerful. But you could extend it to this problem. If God is so powerful can he make a one-way function even he can't reverse? So do one-way functions exist? Sounds like chop-logic and apologia to me And a real non-believer wouldn't say that - because God is a supernatural entity with no proof anyway so why introduce other complications. Personally I think it is a strange question - and you are not the first person to have mentioned it; the question strikes me as incredible hubris. We postulate a supernatural being who by very existence/definition must transcend human bounds and understanding; yet immediately we bring in human constraints and frailties. But this is not the place On 5/19/2017 at 7:22 PM, Trurl said: I believe as humans we are limited and one-way functions do exist. Knowing only N and finding 2 unknowns, knowing only a slight process in which N was encoded is a difficult if not impossible task... It is not merely that humans are limited - it is that maths is axiomatic. We make the foundations of our maths and build from there; within those axioms we can say what is true (sometimes), what is false (sometimes), and what we cannot decide upon (sometimes). Whilst some things are not decidable; if we can prove it within the system of axioms then it is proven and nothing can change that On 5/19/2017 at 7:22 PM, Trurl said: Anyways thanks for the comments. I don't always like being wrong, but I think it is more about being realistic with this problem. I don't know. Would you be interested in seeing a 5 step pattern in multiplication? No one likes being wrong - that's why we study and learn. You are right about doses of realism; you are tackling Mount Everest and getting angry with yourself that a short walk every evening hasn't been enough training. This is an stubborn peak of mathematics that the greatest thinkers have pushed themselves to the limit in an effort to make that vital breakthrough. You are using tools that have been thoroughly tested by people like Euclid, Fermat, and Leibnitz; you can be virtually certain there are no simple things that will crack prime numbers. That is the benefit of fora - post stuff when you have time; someone will look at and critique your 5 step pattern. There are numerous posts in this subforum about the patterns of multiples which are ruled out from being prime Link to comment Share on other sites More sharing options...

Trurl Posted May 23, 2017 Author Share Posted May 23, 2017 I think this problem needs a rest for awhile. I just wanted to post my CAD drawings. They come straight from the Wiki-leaks exploits and are For Internal Use Only. Link to comment Share on other sites More sharing options...

Trurl Posted July 3, 2017 Author Share Posted July 3, 2017 These triangles are starting to look like a polygon. If this interest you, you can view my homepage…at the bottom graphic, click on it for a previous attempt I made back in 2010. Ignore it for now. I just thought a construction exists that will solve for segment FE. Maybe similar triangles between FEF’ and CEC’??? I don’t know. Forget about triangle syx ‘s relationship to Prime products. Concentrate on solving triangle FEC. If it can be solved, a one-way function has just been solved. Link to comment Share on other sites More sharing options...

Trurl Posted July 9, 2017 Author Share Posted July 9, 2017 All right thread viewers, I know you probably don’t see much in my work with Prime numbers. But I assure you it was a sincere effort. What I ask is you look at this link: And find if there is any math problem of value. It is a mixture of work I have done from 2003 till now. Not all of it is about Prime numbers or RSA. Some is based on topics from school. What I need to know is would any of this be of any value for a portfolio? Remember not all my topics are as impossible as Prime numbers. And I would teach those interested in math more traditional subjects. I would not teach them math of the impossible. Though, I would not teach it, I would still post on message boards to those who would want to read. But even though my posts are just ideas, I would teach the curriculum instead. But I still believe that working on the impossible is very important. But it is extra and not the testable program objects that are measured on tests. http://www.constructorscorner.net/ideas_and_gadgets/math/math_home.htm So you may be asking yourself why I try the impossible. Well it was an idea that started with the possible. If you read-through my math write-ups you will see the concepts were simple. But this Prime number thread you are reading now, has made a full circle of all my work. There are many relationships between my standard work and my impossible work. Keep that in mind as you review my math portfolio. And I still have 2 math write-ups that relate to this thread you are reading now. I am trying to keep it realistic and logical. But until then visit the above link. Link to comment Share on other sites More sharing options...

Trurl Posted July 30, 2017 Author Share Posted July 30, 2017 The multiple of 2 "unknown" Prime numbers: 85 = 5 * 17 1/(85 * 3) = 1/(0.0039215) = 255 255/(85 / 3) = 255/(28.333333) = 9 Sqrt[9] = 3 1/(85 * 5) = 1/(0.00235) = 425 425/(85 / 5) = 425/(17) = 25 Sqrt[25] = 5 17^2 = 289 425 - 289 = 136 136 / 17 = 8 17 / 8 = 2.125 1/2.125 = 0.470588 0.470588 * 136 = 64 Sqrt[64] = 8 17/136 = 0.125 8/0.125 = 64 Sqrt[64] = 8 The last 2 calculations are loops. But it shows one thing. These calculations will always be a perfect square root. Thus finding the product of 2 Prime numbers is equivalent to this equation being used to find a perfect square. Yes other products will also have a perfect square, but it will be a decimal and not a whole number. So the idea is to find perfect roots below the product of 2 Prime numbers. We can easily find square roots and we know that this square root is less than the square root of the product. So the numbers we are looking for square root is less than the square root of 85 in this example. 20110818 These are just some interesting relationships I found in between doing class work. I do not know of any equation that finds whole numbers faster than division.The benefit is to find where the mod equals zero.These relationships can be subtracted to each other and set to zero in an attempt to find some meaningful equation. www.constructorscorner.net 20110527--- 20110818 The multiple of two "unknown" Prime numbers. 3.79424E6 = 2459 * 1543 and two non-Prime mutilples 3.79424E6 = 2132 * 1779.66 1/(3.79424E6 * 1543) = 1 / 1.70808E-10 = 5.854551E9 5.854551E9 - (2459^2) = 5.85451E9 5.85451E9 / 2459 = 2.37839E6 The last step is actually to find the modulus of the two numbers. Testing to see if 3 and 28.3333 are Prime multiples of 85: 1/(85 * 3) = 1/(0.0039215) = 255 28.3333^2 = 802.778 255 - 802.778 = -547.778 -547.778/ 28.3333 = -19.3334 28.3333 / 19.3334= -1.46551 1/-1.46551 = -0.682355 -0.682355 * -547.778 = 373.779 Sqrt[373.779] = 19.3334 28.3333 / -547.778 = -0.51724 19.3334/-0.51724 = -373.791 Sqrt[373.791] =19.3337 17^2 = 289 425 - 289 = 136 136 / 17 = 8 17 / 8 = 2.125 1/2.125 = 0.470588 0.470588 * 136 = 64 Sqrt[64] = 8 8/0.125 = 64 The preceding is my work from 2011. It proves nothing, but is an attempt to find simple patterns in Prime factorization. The trouble is that I now have hundreds of pages of Prime product patterns. What is the best way to assemble and organize my work so that it doesn’t get lost. I am not claiming all the patterns work or even are useful, but seeing them in one place may cause someone to have a brainstorm that actually solves something. I have probably a thousand pages of math work. Some works; some impossible; some practical. I have used my website to share are record some of them. And as I have done at SFN, I have sought input and direction. So I have hundreds of files that I don’t think I have time to organize. What is the best format or way to share the files? I thought about sharing everything, but much is incomplete or needs me to explain what I am trying to do. Has anyone ran into the same problem where good ideas have been consumed by too much data created? I saw on 60 minutes that Watson could find better medical therapies than a team of 30 doctors could in a lifetime. Is there no way for humans to decipher data? I know this is no breakthrough. It is just testing for patterns. WholeNumbersIdentifier_unknown_relationship_web_revisted.nb Link to comment Share on other sites More sharing options...

imatfaal Posted July 31, 2017 Share Posted July 31, 2017 I didnt read past the fist set of calcs but even with errors corrected you are just showing basic maths - I will demonstrate error 1/(85 * 3) = 1/(0.0039215) =255 - this is not true 1/(85*3) is 1/255 what you mean is that 1185∗3=255 Well yes it must as 85*3 = 255 and it is basic that 11x=x we learnt it as the bottom of the bottom is the top so what you are doing is in fact 85∗3853 The three on the bottom of the bottom moves to the top 85∗3∗3851 and the 85s cancel each other to leave 3∗3 Link to comment Share on other sites More sharing options...

Trurl Posted August 1, 2017 Author Share Posted August 1, 2017 You should read further. I am aware of the simple pattern that cancels each other out. I canceled many of equations looking for the right one. 1/(85*3) 1/0.0039215686 = 255 255/(85/3) = 9 I know these patterns don’t seem like much, but they were the start of my last equation. The one that you claim will give factors but is slower than division. I argue it gives a position (a distance) from the Prime product. But I am not working on that here I just want to show patterns. I know this isn’t the best written explanation. And I know I have trouble describing my ideas to others. But if you decide to read further down the line there should be a pattern. It has been 6 years since I wrote this. I have to study it myself, because as I stated in my previous post, I have lots of data. Start at the 3rd paragraph of equations. Where 17^2 = 289 for relevant content. Remember = sign means they are equal, but the line below is a pattern where I mixed the numbers together. The below line is not always equal to the top. It is its own operation. This is probably bad practice. But I do not know of a better way to write it down. Link to comment Share on other sites More sharing options...

imatfaal Posted August 1, 2017 Share Posted August 1, 2017 Quote These calculations will always be a perfect square root. Thus finding the product of 2 Prime numbers is equivalent to this equation being used to find a perfect square. Yes other products will also have a perfect square, but it will be a decimal and not a whole number You gave 1/(85 * 5) = 1/(0.00235) = 425 425/(85 / 5) = 425/(17) = 25 Sqrt[25] = 5 1/(1/(x * y))= x * y (x * y)/(x / y) = y(x * y)/x = (xy^2)/x = y^2 Sqrt[y^2] = y What ever whole numbers you use for x and y you will always always always get a perfect square. In fact as the x's cancel then they are immaterial and this would work with x=pi Not gonna read further till you explain your paragraph I have quoted and why it is not completely wrong Link to comment Share on other sites More sharing options...

Trurl Posted August 2, 2017 Author Share Posted August 2, 2017 You are correct. That paragraph is garbage. I am not sure what I meant at the time I wrote it. These calculations will always be a perfect square root. Thus finding the product of 2 Prime numbers is equivalent to this equation being used to find a perfect square. Yes other products will also have a perfect square, but it will be a decimal and not a whole number These calculations will always be a perfect square root. I mean the patterns I just showed are always a square root. Which there is no arguments about. It may seem this is of no use, but I am only showing a pattern; A series of patterns that give use a feel for what is happening in the factorization. Thus finding the product of 2 Prime numbers is equivalent to this equation being used to find a perfect square. In the succeeding patterns of the first write-up only 85/7 equals a whole number. I know this again does not seem to prove anything, but here I am only listing patterns. I will give a better explain my example. Testing to see if 3 and 28.3333 are Prime multiples of 85: 1/(85 * 3) = 1/(0.0039215) = 255 28.3333^2 = 802.778 255 - 802.778 = -547.778 -547.778/ 28.3333 = -19.3334 28.3333 / 19.3334= -1.46551 1/-1.46551 = -0.682355 -0.682355 * -547.778 = 373.779 Sqrt[373.779] = 19.3334 28.3333 / -547.778 = -0.51724 19.3334/-0.51724 = -373.791 Sqrt[373.791] =19.3337 This is true that 19.3337 is not a whole number. Obviously. I know that 5 * 17 = 85 and I will square 17 in part of the following example. Obviously if I know this it doesn’t help me solving knowing only 85. I am only trying to find out what is happening in the factorization. But remember that: 1/(85 * 3) = 1/(0.0039215) = 255 255/(85 / 3) = 255/(28.333333) = 9 Sqrt[9] = 3 But it is interesting that 9 + 19.3337 approximately = 28.3333 It must be tested for all numbers and different potential Prime factors but know: 17^2 = 289 425 - 289 = 136 136 / 17 = 8 17 / 8 = 2.125 1/2.125 = 0.470588 0.470588 * 136 = 64 Sqrt[64] = 8 8/0.125 = 64 Is unique among other factors. It is the correct factor. This pattern does not solve that, but shows something interesting is going on. We must test for more for more numbers, but the same is true for 85/11. I am not solving a pattern in Prime numbers. I only care what happens when factored. I know it doesn’t seem like much but it is how I visual the equations. For example the factors of 85 have to be less than 85. And when you choose the smaller factor the second largest factor is limited in value. I know it has been tried and there are already algorithms. But I just want to show that my idea is unique. And I realize it uses test values. But these patterns are important in understanding how I derived my equation. The one that is ugly and slower than division. But again, what patterns are going on in the calculation of that equation. That is why I shared this. To reiterate: 85 *11 =935 935/(85/11) = 121 Sqrt(121) = 11 85/11 = 7.72727 Abs[(7.72727^2 - 935] = 875.293471 875.293471/7.722727 = 113.27322731 Abs[113.27322731 -121] = 7.7267 The solution of 7.7227 is similar to what happen with 85/3 as 28.333. The square 9 for 3 and square 121 for 11 shows a pattern in the factoring. That is if I just didn’t redundantly cause the pattern. I hope it is understood what I tried to present. Link to comment Share on other sites More sharing options...

imatfaal Posted August 4, 2017 Share Posted August 4, 2017 LT;DR But to reiterate if y is whole number then answer will be whole number 1/(1/(x * y))= x * y (x * y)/(x / y) = y(x * y)/x = (xy^2)/x = y^2 Sqrt[y^2] = y this is what you are doing. So please look at the simple algebra and realise what ever number you put in for y you will always get y out again. NB 1. Perfect squares means whole numbers - you cannot have a perfect square which when rooted gives fractional answer by definition 2. Factors are also whole numbers - there is little point looking for divisors with fractions as you will always always get one Link to comment Share on other sites More sharing options...

Trurl Posted August 13, 2017 Author Share Posted August 13, 2017 Yes, imatfaal is correct those patterns are true for all numbers. I intend to redeem the mistake. Early in my work I was just finding different ways to represent “y” in terms of “x”. It led to a lot of x = x. But I did have several equations that were distinct and significant. Below are some patterns that are redundant, but I encourage you to look because if I do find one that is distinct it is gold dust. I believe the equation at the start of this post to be distinct. I know it cannot be solved. But I think it is a pattern. Yes, my equation solved N knowing x, which wasn’t significant. But if there a way to solve the equation N would be the only number needed. So here we go again: PNP= 85 x=5 {(((((PNP^4)/x)+2*(PNP^2*x^2)+x^5)/PNP^3)* x^3) / PNP } 85 5 {2160900/83521} N[{2160900/83521}] {25.872535051065`} Sqrt[25.8725] 5.0865 PNP= 85 x=5 {((PNP^4+2*(PNP^2*x^2)+x^5)/PNP^3)} 85 5 {420520/4913} N[{420520/4913}] {85.5933} N[{2102600/4913}] {427.967} 427.966619173621`/85 5.0349 ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------___ ___ ___ ___ ___ ___ ___ _ Other Patterns related to above work 125/5 = 25 25* 52200625 = 1305015625 --------------------------- 52200625/614125 =N ___________________________________ 52200625 / 125 = 420520 _______ 85^3 = 614125 614125 / 125 = 17^3 = 4913 ________ 83521/ 4913 = 17^4 / 17^3 = 17 _____________________ 2102600 / 420520 = 5 MaybeSIGSFN20170813V02.nb Link to comment Share on other sites More sharing options...

Trurl Posted September 16, 2017 Author Share Posted September 16, 2017 PNP = 85 (((((x^2*PNP^4 + 2*PNP^2*x^5) + x^8)/PNP^4) - ((1 - x^2/(2*PNP))))*((PNP^2/ x^2))) == PNP^2 85 (7225 (-1 + x^2/170 + (52200625 x^2 + 14450 x^5 + x^8)/52200625))/x^2 == 7225 Solve[(7225 (-1 + x^2/170 + (52200625 x^2 + 14450 x^5 + x^8)/52200625))/x^2 == 7225, {x}] {{x -> Root[-104401250 + 614125 #1^2 + 28900 #1^5 + 2 #1^8 &, 1]}, {x -> Root[-104401250 + 614125 #1^2 + 28900 #1^5 + 2 #1^8 &, 2]}, {x -> Root[-104401250 + 614125 #1^2 + 28900 #1^5 + 2 #1^8 &, 3]}, {x -> Root[-104401250 + 614125 #1^2 + 28900 #1^5 + 2 #1^8 &, 4]}, {x -> Root[-104401250 + 614125 #1^2 + 28900 #1^5 + 2 #1^8 &, 5]}, {x -> Root[-104401250 + 614125 #1^2 + 28900 #1^5 + 2 #1^8 &, 6]}, {x -> Root[-104401250 + 614125 #1^2 + 28900 #1^5 + 2 #1^8 &, 7]}, {x -> Root[-104401250 + 614125 #1^2 + 28900 #1^5 + 2 #1^8 &, 8]}} N[%3] {{x -> -24.3484}, {x -> 4.97889}, {x -> -4.2072 - 2.87925 I}, {x -> -4.2072 + 2.87925 I}, {x -> 1.71775 - 5.00069 I}, {x -> 1.71775 + 5.00069 I}, {x -> 12.1742 - 21.0804 I}, {x -> 12.1742 + 21.0804 I}} f[x_] := (((((x^2*PNP^4 + 2*PNP^2*x^5) + x^8)/ PNP^4) - ((1 - x^2/(2*PNP))))*((PNP^2/x^2))) == PNP^2 f'[x] (7225 (x/85 + (104401250 x + 72250 x^4 + 8 x^7)/52200625))/x^2 - ( 14450 (-1 + x^2/170 + (52200625 x^2 + 14450 x^5 + x^8)/52200625))/x^3 == 0 g[x_] := f'[x] 1 + (85 - f[1])/f'[1] False 313875685/208889212 N[313875685/208889212] 1.50259 Again the above equation has 5 as a solution knowing only N. I know this is not enough until tried with all values, but it seems that Mathematica has a solution for this instance. I want to show what I was trying to do at the bottom (end) of the equations. I was trying to use Newton's Method. I studied the equations from Wikipedia and felt that maybe applying the method would simplify my equations. I used an y of 85 (a y of N, instead of zero) and placed the slope (the derivative of my PNP equation) in order to solve a test value of x. This test value x is intended to be an estimate, however I tried to find a modified the equations to find x based on a given number (start point; 1) and the slope (derivative) of the original equation (from the first post of this thread). I know this didn't work, but I share just to walk through the idea. If there was a way to find a given value of an equation using the equation and its slope, it would solve my original equation. Below is the equation. 0 = f ' (xn) *(x - xn) + f(xn) 85 = f ' (xn) *(x - xn) + f(xn) xn = 1 ---- This is the start value. 85 = f ' (xn) (x - xn) + f (xn) 1 + [85 - f(xn) / f '(xn)] = x I am aware this doesn't work. However I thought the idea was so simplistic it could work. This is what I am proposing: The slope is known and any start number can be used, so finding the value of x where the y-coordinate equals N can be found without the complexity of the original equation. Obviously, it has been done before. But I am asking for help in trying to apply it to my equation. 20170916SFNnewtonMethod008b.nb Link to comment Share on other sites More sharing options...

Trurl Posted September 25, 2017 Author Share Posted September 25, 2017 PNP = 85 (((((x^2*PNP^4 + 2*PNP^2*x^5) + x^8)/PNP^4) - ((1 - x^2/(2*PNP))))*((PNP^2/ x^2))) == PNP^2 85 (7225 (-1 + x^2/170 + (52200625 x^2 + 14450 x^5 + x^8)/52200625))/x^2 == 7225 Solve[(7225 (-1 + x^2/170 + (52200625 x^2 + 14450 x^5 + x^8)/52200625))/x^2 == 7225, {x}] {{x -> Root[-104401250 + 614125 #1^2 + 28900 #1^5 + 2 #1^8 &, 1]}, {x -> Root[-104401250 + 614125 #1^2 + 28900 #1^5 + 2 #1^8 &, 2]}, {x -> Root[-104401250 + 614125 #1^2 + 28900 #1^5 + 2 #1^8 &, 3]}, {x -> Root[-104401250 + 614125 #1^2 + 28900 #1^5 + 2 #1^8 &, 4]}, {x -> Root[-104401250 + 614125 #1^2 + 28900 #1^5 + 2 #1^8 &, 5]}, {x -> Root[-104401250 + 614125 #1^2 + 28900 #1^5 + 2 #1^8 &, 6]}, {x -> Root[-104401250 + 614125 #1^2 + 28900 #1^5 + 2 #1^8 &, 7]}, {x -> Root[-104401250 + 614125 #1^2 + 28900 #1^5 + 2 #1^8 &, 8]}} N[%3] {{x -> -24.3484}, {x -> 4.97889}, {x -> -4.2072 - 2.87925 I}, {x -> -4.2072 + 2.87925 I}, {x -> 1.71775 - 5.00069 I}, {x -> 1.71775 + 5.00069 I}, {x -> 12.1742 - 21.0804 I}, {x -> 12.1742 + 21.0804 I}} f[x_] := (((((x^2*PNP^4 + 2*PNP^2*x^5) + x^8)/ PNP^4) - ((1 - x^2/(2*PNP))))*((PNP^2/x^2))) == PNP^2 b = [Integrate[f, {x, PNP - 1, PNP}]] b = f[85] The above equation is simplifying the PNP polynomial equation from SFN post 1. This equation proves to be true but there is no way to solve it. The equation shows where x is as x approaches N (or PNP). Taken this feature and using the integral to simplify the original equation where PNP is 85 (in this example), we know that the integral from N - 1 (that is 85 - 1) to N (which is 85) is equal to f(85). Of course some figuring is wrong here. I haven't taken a calculus class in almost 20 years. I don't know if it is possible to have the integral equal the original function. But intuitively it seems to work. I know fault will be found here. I'm just bouncing off ideas. I would have formatted this post better, but Big Bang Theory is coming on. Link to comment Share on other sites More sharing options...

Trurl Posted October 4, 2017 Author Share Posted October 4, 2017 Attached are 2 important PDF’s that have been on my site for years, but are probably only seen by a few because of the difficulty of going through all the information. Yes, some patterns may repeat. But here in the first PDF, 3 equations are presented. Pay close attention to: Sqrt[(N*y – x^2)/x] = y This is one pattern for sure that I know isn’t just repetition. It is used in the equation at the beginning of this discussion here at SFN. The other 2 equations are just hypothesis. You will see this in the second PDF: I believe with most symmetric key ciphers, the fact that Prime numbers equal a one way function does not mean that there isn’t a pattern in the one way computation. So if N equals the product of 2 Prime numbers, Prime numbers have no pattern, but multiple 2 Prime numbers together there is a pattern. Where x = 571 and y = 1381 1381 / 2p = 219.7929764 2p * (remainder(1381/2p) * 2p) * 571 = 13208.65186 rewritten: 2p * ((1381/2p-whole number part) * 2p) * 571 = 13208.65186 13208.65186 / (1381/2p) = 60.09 13208.65186 / 60.09 = 219.7929764 219 = 219 And another relationship: New equations: 2p * (remainder (17 / 2p)) = 4.42 4.42 * 5 = 22.1 22.1 * (5 / (17* 2p)) = 0.884 0.884 * 5 = 4.42 4.42 = 4.42 See what you can do with it and share it. I believe p stands for Pi. Some of these write-ups are several years old and after making so many I don’t always know what I intended to communicate, without studying them. But I feel these PDF’s are important. That is important enough these write-ups may give meaning to the work. Either way, let me know if you think they are trash, or if it gives you any ideas. 20140519TrigPrimes005secCopyCC.pdf PrimeProductSolutionFlyerCopyCC.pdf Link to comment Share on other sites More sharing options...

Trurl Posted November 6, 2017 Author Share Posted November 6, 2017 I know this doesn’t solve the problem. I am only trying to solve the triangle I constructed. But is it possible to find FE by subtraction. I may not be doing it right, but is it possible to do subtraction of the triangle segments to get FE? To me it seems possible, but when you go to do it, it is confusing. But it just feels possible. Remember I am only trying to solve for FE. AC = N [Absolute value [ AC – AE – (AC-CE) – CE]] = FE Link to comment Share on other sites More sharing options...

Trurl Posted December 25, 2017 Author Share Posted December 25, 2017 Merry Christmas everyone! My present for you is a problem that was given to me by a friend Curtis Blanco. He proposed that I try and solve this geometry problem. I thought it would be best suited for a geometric construction. If you follow this link I have written it up 10 years ago. Here is the problem: Two buildings, I and II stand next to each other forming an alleyway between them. Two ladders, ladder A and ladder B in the alley cross each other touching at the point where they cross. The bottom of A sits against the base of building I, and leans over on building II. The bottom of ladder B sits against the base of building II, and leans over on building I. Ladder A is 3 meters long, ladder B is 4 meters long. The point where ladder A and ladder B cross is 1 meter above the ground. What is the width of the alleyway? http://www.constructorscorner.net/ideas_and_gadgets/math/math_hunch/hunch_00001/hunches_section0003_fellow_constructors/ladder_circles.htm It took me a while to figure out what I was trying to do back then, but I think it works. I bring this up because it now relates to my Prime problem. In the Travelling Salesman problem, I proposed using circles to find distances. When multiple circles intersect they give clues on what is the shortest path between points. The difficulty of the TSP is not finding the shortest line distance between points. Instead finding the point this way doesn’t always lead to the shortest distance, if the pattern is in a square for instance. The perimeter of the square is longer than navigating through the center. All of this learned from Wikipedia. I still need to research the problem. I show this here because I think it is interesting and plan to relate it to my overall problem of Primes. I know I have to make a better diagram. I am behind on drawing these diagrams. My job title was once “document specialist”. I know I should be better with this, but version updates in software packages, has made my old programs incompatible with Windows 64 bit. But follow the link and see if it solves the problem I listed. I will follow up to this post with more descriptive work, because I know how difficult it is to follow my description of the solution. But I hope it shows that with the TSP, circles are our friends. Link to comment Share on other sites More sharing options...

Trurl Posted January 1, 2018 Author Share Posted January 1, 2018 Ok for the new year I wanted to clarify the equation I posted here. With the following equations I wanted to show the test value of x produces a PNP – the calculated PNP approaches zero that test values of x equal p, in N = p* q. So, when F approaches PNP the value of x is the p, in N = p * q. I have included 2 if-statements that test this. And as you can see, an x equal to PNP (85 in this case) is zero, but those numbers larger than the correct x of 5, increase in value as they approach 5 and those test values smaller than 5, increase until they reach zero. (That is in my IF-Statements.) Yes, I know that the test value is too small. The problem is the accuracy of the PNP calculation relies on the square root of a large value. This is causing a margin of error in values of PNP greater than 3 digits. But I show this because the estimate is significant. How do I make the square root of the polynomial in F in this code to be more accurate? I hope you agree this equation is significant. PNP = 85 x = 85 F = Sqrt[(((((x^2*PNP^4 + 2*PNP^2*x^5) + x^8)/ PNP^4) - ((1 - x^2/(2*PNP))))*((PNP^2/x^2)))] If [PNP - F > 0, x = x + Sqrt[PNP - F + x]] If [PNP - F < 0, x = ( x - Sqrt[F - PNP + x]) /2 ] 85 85 161 Sqrt[4123/2] 1/2 (85 - 7^(3/4) Sqrt[23] (589/2)^(1/4)) N[1/2 (85 - 7^(3/4) Sqrt[23] (589/2)^(1/4))] -0.249277 PNP = 85 x = 5 F = Sqrt[(((((x^2*PNP^4 + 2*PNP^2*x^5) + x^8)/ PNP^4) - ((1 - x^2/(2*PNP))))*((PNP^2/x^2)))] If [PNP - F > 0, x = x + Sqrt[PNP - F + x ]] If [PNP - F < 0, x = ( x - Sqrt[F - PNP + x]) /2 ] 85 5 Sqrt[4179323/2]/17 1/2 (5 - Sqrt[-80 + Sqrt[4179323/2]/17]) N[1/2 (5 - Sqrt[-80 + Sqrt[4179323/2]/17])] 1.37825 PNP = 85 x = 7 F = Sqrt[(((((x^2*PNP^4 + 2*PNP^2*x^5) + x^8)/ PNP^4) - ((1 - x^2/(2*PNP))))*((PNP^2/x^2)))] If [PNP - F > 0, x = x + Sqrt[PNP - F + x ]] If [PNP - F < 0, x = ( x - Sqrt[F - PNP + x]) /2 ] 85 7 (11 Sqrt[45773587/2])/595 1/2 (7 - Sqrt[-78 + (11 Sqrt[45773587/2])/595]) N[1/2 (7 - Sqrt[-78 + (11 Sqrt[45773587/2])/595])] 1.88414 PNP = 85 x = 3 F = Sqrt[(((((x^2*PNP^4 + 2*PNP^2*x^5) + x^8)/ PNP^4) - ((1 - x^2/(2*PNP))))*((PNP^2/x^2)))] If [PNP - F > 0, x = x + Sqrt[PNP - F + x ]] If [PNP - F < 0, x = ( x - Sqrt[F - PNP + x]) /2 ] 85 3 Sqrt[847772947/2]/255 3 + Sqrt[88 - Sqrt[847772947/2]/255] N[3 + Sqrt[88 - Sqrt[847772947/2]/255]] 5.69458 PNP = 85 x = 1 F = Sqrt[(((((x^2*PNP^4 + 2*PNP^2*x^5) + x^8)/ PNP^4) - ((1 - x^2/(2*PNP))))*((PNP^2/x^2)))] If [PNP - F > 0, x = x + Sqrt[PNP - F + x ]] If [PNP - F < 0, x = ( x - Sqrt[F - PNP + x]) /2 ] 85 1 (7 Sqrt[13123/2])/85 1 + Sqrt[86 - (7 Sqrt[13123/2])/85] N[1 + Sqrt[86 - (7 Sqrt[13123/2])/85]] 9.90669 Above is the input and output of my code. The test values are separated by spaces. Link to comment Share on other sites More sharing options...

Trurl Posted January 26, 2018 Author Share Posted January 26, 2018 Taking a break on rather you agree or disagree that my equation has any value to the products of semi-primes. Back in 2003, I was with a group of friends whom were killing time. There was talk about a simple check-book-balancing sheet. However, the discussion was not yet about math and we were not doing school work. But we found ourselves in a common room that had a clock. Somehow someone in our small group noticed this clock was different from his wrist-watch. After everyone became interested, someone in the group ask if both clocks continued to run how much would on clock have to lose each hour to eventually have the same time as the other clock. Yes, I now know we were discussing the modulus. However, there is a twist here. The test to see if the times will ever be the same is testing to see if the times are relatively Prime to each other. Also, if you pick the time lost in order to synchronize the clocks, you may actually increase the difficulty of the problem. But what if you add more clocks or use smaller time increments such as seconds and fractions of seconds. I know this problem, though original to our group, has been thought of before. But this problem is an extension of the common modulus represented by a clock. This may be significant to the semi-Prime products, because the problem of many small clocks is the same one to find products. Let me know what you think. I wrote this, and I think it has some solid thought. It has been awhile since we had this idea and I hope a recall it factually. The important part is that the idea stuck with me. I tried to keep the description short here. Adjusting one clock will affect how the other clocks need adjustment. I think the multiple-clock-description fits nicely with cryptography. Link to comment Share on other sites More sharing options...

Trurl Posted February 9, 2018 Author Share Posted February 9, 2018 This is a question to the mathematicians on the forum: when you're trying to solve an open problem in pure mathematics, what are the first things you do? Do you test the conjecture with a few example problems? Do you look up recent theorems related to the question, or do you just dive right in? I wanted to answer your question without killing your thread and leave it to professional mathematicians to answer, so I answered here. I do math for fun. And I choose my topics simply by what interests me. I use an intuitive method. I try to picture the problem completely and think if I have any techniques that will take me in the direction I feel that will solve the problem. But most likely will revel other problems and become a learning experience. Now with the Internet, math research is easily accessed. This is great to research, but it often leads to an overwhelming amount of information or confusion when piecing together conflicting evidence. This is why finding your “own” problem you want to work on is difficult, because the start point is a judgement call. But nothing helps with math problems than just searching and solving as many as you can. The process comes with practice, because you must develop a personalized technique. I have read books on “Flow” where a psychologist is trying to figure what makes someone creative. I think it helps to know how they approach a problem and though processes they go through. But I also think you can be consumed by someone else’s methods while it is more important to develop your own. I will give you an example on a problem I want to research. Everyone is mining Ethereum, the most popular digital currency. I have been working on an algorithm that will test values to see if they are factors. (Yes, I know it needs improved.) So, the first thing I look at is the enciphering protocol that Ethereum uses for its contracts. I have a little understanding of RSA, but a search reveals AES is used. I look at AES and more reading reveals that this encryption standard is well proven and nothing to do with factoring. But more reading on Wikipedia gives me a layman’s explanation on how substitution and movement in many iterations across matrices. Ok so I have a little knowledge of matrices from linear algebra. And I know a public key is used to encipher the message. But there is no way I can solve the pattern the computer enciphers with. But I look to what I know. We often use matrices to solve vectors. And if I took plain text and enciphered it with the known, public key it may show more than expected. If I could somehow give values to the plain-text and treat it as a vector addition, isn’t the public key the resultant of the matrices. I know there are a lot of unknowns here. And the idea is just a hunch; an intuitive idea. But that doesn’t mean the hunch does not need explored further. But what if a free-body-diagram with the forces being “movements of data” and the result force being the public-key. I know the idea probably won’t work. I just wanted to share an idea and how it can lead to something worthwhile. For this reason, I put so much effort in “The Products of Primes” here in this thread. Link to comment Share on other sites More sharing options...

Trurl Posted February 13, 2018 Author Share Posted February 13, 2018 In[50]:= PNP = 85 x = 5 F = Sqrt[(((((x^2*PNP^4 + 2*PNP^2*x^5) + x^8)/ PNP^4) - ((1 - x^2/(2*PNP))))*((PNP^2/x^2)))] Out[50]= 85 Out[51]= 5 Out[52]= Sqrt[4179323/2]/17 In[53]:= N[F] Out[53]= 85.0333 Ok, the important thing about the above equations is equation F is the Cumulative Distribution Function of the factors of PNP! The values below are to show that with the proper x, F will equal (within small error) PNP. This is just to show that it works for Semi-Primes that are a little harder to do on a calculator. If someone knows how to program large numbers: millions of digits, then they could find larger Prime numbers or break encryption that use factorization as a one way function; RSA for example. But wait. If you put F into the following programming logic you have created a Normal Distribution Function! If [PNP - F > 0, x = x + Sqrt[PNP - F + x]] If [PNP - F < 0, x = ( x - Sqrt[F - PNP + x]) /2 ] These logic statements create a normal distribution when graphed. Of course it is centered at zero and can be used to correct the error of equation F. But we must note it is mirrored in the x-axis. But know we know where the distribution of the smaller factor that is multiplied to equal PNP. We have a Bell Curve; almost. I am only calling it that because that is what is usually though of with normal distribution. But I think as you read this it may add more credibility to my work. In[41]:= PNP = 2999* 6883 x = 2999 F = Sqrt[(((((x^2*PNP^4 + 2*PNP^2*x^5) + x^8)/ PNP^4) - ((1 - x^2/(2*PNP))))*((PNP^2/x^2)))] Out[41]= 20642117 Out[42]= 2999 Out[43]= Sqrt[40378385476407827918859/2]/6883 In[44]:= N[F] Out[44]= 2.06434*10^7 20180213NormalDistribution7PM02.nb Link to comment Share on other sites More sharing options...

Trurl Posted February 23, 2018 Author Share Posted February 23, 2018 Ok, this is my final post to this thread, unless someone asks a question. I will continue to work on this problem. And I thank the community of SFN for letting me share my math problem, even when it sounded impossible. I wish more would have commented. As of this post it has been viewed 10,500 times. That is a significant amount. I am also asking permission to use my post in other writings. I know the forum is free, but is it still permissible to use those posts of members who ask questions? I am using them only so my posts make sense, in the proper context. Anyway, I hope you have enjoyed the work I put in on these posts. And I hope I made you believe that reversing the N = p * q problem is possible. But often in math, the idea is just as important as the solution. If you did not believe me these equations would solve anything, perhaps you considered it for one moment. Now that it has concluded, feel free to post any of your thoughts. I thought this problem would lead to great conversations. Don’t let a math thread get less response than one on astrology. Link to comment Share on other sites More sharing options...

Trurl Posted July 24, 2018 Author Share Posted July 24, 2018 As you know I have a complex equation describing semi-Primes. The fact is I cannot at this time solve the polynomial. But instead, it is more of a guessing attack to discover Prime factors knowing only N. What I have discovered is more of a series. It only works with Prime factors. More importantly, the series works with the same Prime factor multiplied by other Prime factors. So: · The series is a guess attack, comparing N to computed N. · The series only works for semi-Prime factors that are Prime numbers. · The series works for a given Prime number and infinite other, different, Prime number factors, multiplied for different results of different N’s. (If the equation did not work for multiple factors, multiplied by the same factor, the equation would be useless.) · The equation that produces the series can be used to test if a number is truly Prime. Other significant facts of the equations: · The equation is too complex and factors with imaginary numbers. · The equation is simple algebra. · Even though the equation cannot be solved with a perfect mathematical answer, the series is beneficial to computer algorithms. · The equations may have relationships to logarithmic spirals. I have been working with Prime numbers and Logarithmic spirals since 2006. But have recently tried to switch my efforts to other projects. I haven’t posted to SFN for 5 months, in this thread. But instead of trying to defeat RSA and use my math series to factor p and q knowing only N; there is a twist. If you know p (also what I usually refer to as x), you can test to see if it is a Prime number by multiplying by Prime number test values. If the equations hold true, then a test for Primality exists. https://www.3dbuzz.com/forum/threads/200441-New-One-Way-Function I don’t know if anyone believes me when I say this series is significant. But I posted the link to me “old” work from the time before I found the more useful equations. Much of that work is just plain wrong. But it will show you how I ended up with the equations I promote now. So, if you think there is any meaning to my equations, please post me a message in this thread to let me know. I will respond with more information. Link to comment Share on other sites More sharing options...

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