Strange 2645 Report post 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/ Upvote Downvote Like × Quote Share this post Link to post Share on other sites

imatfaal 2479 Report post 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 Upvote Downvote Like × Quote Hide imatfaal's signature Hide all signatures A little learning is a dangerous thing; drink deep, or taste not the Pierian spring:there shallow draughts intoxicate the brain, and drinking largely sobers us again.- Alexander Pope feel free to click the green arrow ----> Share this post Link to post Share on other sites

Trurl 12 Report post 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? Quote Share this post Link to post Share on other sites

imatfaal 2479 Report post 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 Upvote Downvote Like × Quote Hide imatfaal's signature Hide all signatures A little learning is a dangerous thing; drink deep, or taste not the Pierian spring:there shallow draughts intoxicate the brain, and drinking largely sobers us again.- Alexander Pope feel free to click the green arrow ----> Share this post Link to post Share on other sites

Trurl 12 Report post 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. Quote Share this post Link to post Share on other sites

Trurl 12 Report post 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. Quote Share this post Link to post Share on other sites

Trurl 12 Report post 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. Quote Share this post Link to post Share on other sites

Trurl 12 Report post 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 Quote Share this post Link to post Share on other sites

imatfaal 2479 Report post 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 Upvote Downvote Like × Quote Hide imatfaal's signature Hide all signatures A little learning is a dangerous thing; drink deep, or taste not the Pierian spring:there shallow draughts intoxicate the brain, and drinking largely sobers us again.- Alexander Pope feel free to click the green arrow ----> Share this post Link to post Share on other sites

Trurl 12 Report post 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. Quote Share this post Link to post Share on other sites

imatfaal 2479 Report post 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 Upvote Downvote Like × Quote Hide imatfaal's signature Hide all signatures A little learning is a dangerous thing; drink deep, or taste not the Pierian spring:there shallow draughts intoxicate the brain, and drinking largely sobers us again.- Alexander Pope feel free to click the green arrow ----> Share this post Link to post Share on other sites

Trurl 12 Report post 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. Quote Share this post Link to post Share on other sites

imatfaal 2479 Report post 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 Upvote Downvote Like × Quote Hide imatfaal's signature Hide all signatures A little learning is a dangerous thing; drink deep, or taste not the Pierian spring:there shallow draughts intoxicate the brain, and drinking largely sobers us again.- Alexander Pope feel free to click the green arrow ----> Share this post Link to post Share on other sites

Trurl 12 Report post 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 Quote Share this post Link to post Share on other sites

Trurl 12 Report post 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 Quote Share this post Link to post Share on other sites

Trurl 12 Report post 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. Quote Share this post Link to post Share on other sites

Trurl 12 Report post 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 Quote Share this post Link to post Share on other sites

Trurl 12 Report post 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 Quote Share this post Link to post Share on other sites

Trurl 12 Report post 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. Quote Share this post Link to post Share on other sites

Trurl 12 Report post Posted just now 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. Quote Edit Share this post Link to post Share on other sites