Trurl

Hypothesis, Given a triangle with known N, the similar triangle formed at any given triangle with the largest side equal to N, will have a similar triangle with sides equal to p and q. This triangle will have the largest side similar to N. This is probably absolutely wrong, but I base it on the fact that the 2 sides multiplied together equal N. I know I am probably breaking a rule of sine and cosine by addition of trigonometry, but my logic is this: If p and q are the products of N then a triangle the contains N as the largest side, the triangle will have a similar triangle that has sides of the products. After all, a similar triangle is just proportions and since p and q are proportional to the original triangle, multiplication does solve the similar triangle. These similar triangles solve the one-way function of N = p *q. The reason it isn’t easy to visualize is because to find the answer because we did not have the equation to find the relationship between p and q. It is ok if I am wrong. This is after all an impossible one-way-function triangle. But if you read through it, you may understand what I was attempting to do. I will respond to each individual post later. I just wanted to clarify my idea, if possible. Yes, I know there are infinitely many triangles. I’m counting on it so that my sides equal to p and q exist. Again, they always exist. I am just using properties of triangles to simplify an equation that is used to find p and q knowing only N.

I know no one believes in my problem. But I need someone to follow along with the solution (or disprove my ideas). It is something a problem solver must work through themselves. I know FC does not equal CE. And if we can’t solve for FE or find FC. I will not break the rules of geometry by using my side-angle solution. But I propose a math exercise to go through. Instead of believing my solution, prove the exercise wrong. (Probably not that hard to do, but I think it is worth trying.) FE = CD / cos[angle ECD + angle FCE] = CE / cos[FCE] This may be solvable with FE the unknown, but we need to solve for CD and CE to for this to work. Angle FCE is also unknown. N/sin[120] = CE/sin[60] To me I drew this triangle to represent the one-way function. I believe it is solvable. This is why I have reposed it. Proving this triangle solvable means that semi-Primes are no longer able to be used in cryptography. It sounds simple enough, but this is no ordinary triangle. I know when I start talking about Prime number solutions or one-way-functions or solving impossible triangles, it draws a red flag. If it were so easy someone would have solved it already. But if you do not believe we can reach a solution, you are probably correct. It may seem like I do not know math because I am always looking for work-arounds. I know the problem is unsolved and I know the probability of finding a solution equals my probability of solving this triangle. But I believe my approach is different. Solved or unsolved semi-Primes, it is still worth the mathematical exercise.

I could make this long, but I will get to the point. I realize most don’t like my Prime equations. I design these triangles as a graphic representation of my problem. Do you agree that triangle yxs is similar to triangle AFC? Do you believe that triangle AFC has no importance other than the fact that it has PNP (also called N) equals the product of 2 unknow Prime numbers? Do you believe that since N is known and that AFC can be solved for and that triangle yxs angles are the same because of the definition of similar triangles? Most importantly, do you believe that y and x are equal to in proportion to N? (This is the most important step, I am still working on.) If y is the larger Prime factor and x is the smallest, we will use similar triangles y/x = AF/FC. All angles and sides of AFC are known. The proportions of y/x can be written with the following equations: N = 85 y = (N^2/x) + x^2) / N x = x N = y * x, but we only know N y/x = AF/FC AF/FC is known as solved by the triangle and is a real number proportion. So (N^2/x) + x^2) / N * (1/x) = proportion AF/FC Calculate it out I am still working on this. I got x = Sqrt(N/ proportion AF/FC) Again I am still verifying everything is mathematically proper. But I post here to get feedback.

Ok, here is my question simplified: Given: two similar triangles ABC and DEF, Are segment AC divided by segment AB proportionate to segment DF divided by segment DE? If in triangle ABC, all lengths are known, and all angles are known; And in triangle DEF all angles are known; Using the similarity of AC/AB = DF/DE DF and DE are known, but AC and AB are known only by equations, can the equation between AC and AB be simplified by relation to the known DF/DE? That is my question simplified. I did not include example because the problem relies on AC/AB = DF/DE. Other questions I have just realized that the Law of Sines relates to similar triangles. But what I was not taught in school is: d = Sin (a) / A = Sin (b) / B = Sin (c) / C We were not taught that d equals the proportion and is the circumscribed circle of the triangle. But this can be discussed later. I just need to know if what I described above is possible. To me it makes sense and is simple. But I can’t remember ever studying similar triangles this way and I don’t know if a text book would list it under its laws. But I think it isn’t there just like we didn’t learn about “d” because the application is limited. I could be completely wrong. But a side divided by the other in one triangle should equal a side divided by another in a second, similar triangle.

Ok, you are going to laugh at this one. But I am seriously asking this question. I need to think outside the box if I want to break a one-way-function. In this case outside the triangle. My question is: Why can’t I combine the Law of Sines and the property of similar triangles to solve 2 different similar triangles of different sizes? And secondly why can’t I say the proportions in one triangle are not proportional to the same proportion in a similar triangle? I mean if all sides and angles are known in one and the other similar triangle all the angles are known, there has to be a method to solve the segments of the unknown, second similar triangle. Is there a method I do not know of? For example, you know the lengths and angles of triangle one; and you know the angles of a smaller, similar triangle angle two. I know, if you new one side of angle 2 you could solve with similar triangles no problem. If it were that easy the problem wouldn’t be a one-way-function. I just wanted to get advice on the possibility and if anyone sees this as a worthy problem. To be specific I do not know the lengths of the segments on angle 2, but I do know the equations that make up the lengths. Does anyone believe it would be possible to solve angle 2 with only equations as the known for angle 2? I have put much thought into this. Because being able to break a one-way-function is the result. I know that there are infinite similar angles. And yes, I know trigonometry. This has just been bugging me. Why can’t I do this? Or it must be possible? But are there any higher geometry or higher math’s that solve such a problem? I’m sure someone has encountered this problem before.

Ok, I have a thought experiment; a visual model. I used to run daily. I would run down the trail at an 8-minute mile. Somehow a fly that has a maximum speed of 5 mph lands on my neck. I swat the fly and it leaves my neck. Accordingly, I increase my speed to a 6-minute mile. The same fly somehow lands on me again. Every time it leaves my body from my swat in comes right back. The thing to ponder is that the fly moves slower than me yet can intercept me and after intercepting me leaves my body only to intercept me again. What are some physical properties behind this? I would like to get a variety of answers. But to show what I was thinking: suppose drones were set up to intercept a jet aircraft or missile. They don’t have the speed to match the missile or time to acquire exact coordinates. Could a drone intercept a missile as the fly has intercepted me (multiple times)? And would being able to intercept multiple times solve the problem of finding exact coordinates of the missile?

I think I have an understanding of what you are doing with factoring. You just want it in that form. I would like to see an updated proof, since this thread has so many explanations. This is the first time I saw this conjecture, so I hope some experts on this forum take interest. The problem is worth working through. Right or wrong it is a fresh approach. Just because the conjecture is unsolved does not mean the idea isn’t valid.

I think the approach is brilliant and maybe simple enough to work. But I am still not convinced. You take 2 to a power and multiply by Prime factors. I agree that you can build any number this way, but I don’t believe multiplying by 2 here is the same as dividing an even number by 2 in the conjecture. The order of operations. And I cannot test this algorithm in a computer since we cannot factor the numbers in this series. Again I’m probably wrong but this is not a linear series. My understanding is that you are building a pattern linearly and the conjecture is not a one-to-one function. This is just my understanding but if you can prove your method does this I will buy multiple copies when you get it published.

Ok, I’m listening. Where does this go from here? This is how I understand you: Add 1 to a Prime number and it can be written as a power of 2. Since all Composite numbers factors are Prime numbers, those Prime factors added to 1 reduce to a power of 2. This will reduce to 1 and prove the conjecture. Your job is to prove this is possible. My question is how do you factor the Primes and apply the conjecture without changing the value of original number. By this I mean you factor the number into two Prime numbers and add 2 (+1 each Prime) to coverage to 1 and prove the conjecture. But at the same time you did not apply 3x +1 to the original value breaking the series. I am probably wrong again. I just don’t follow the modifications to the conjecture series. I’m am just making clear if I am following what you are doing. But if you explain this, I’m onboard. I am still looking at the original proof to make sure my question made sense. I am not completely sure of all the functions affecting x. But if the above question is confusing I will simply ask this: Are you breaking the rules of the conjecture by apply functions?

Ok here is what I don’t yet understand. Try a series for 85. In your example you would multiply by 3 and add 1. This would find Prime factors for the new number and not 85. I believe that would make it hard to factor Primes or semi-Primes. But then again I don’t have a full understanding of the problem. You are the expert on this problem. This is your problem. I am just making you defend and explain it more clearly. Also I took interest in you explaining your idea. I don’t know if you are right or wrong. A mathematician must decide for himself if a problem is worth pursing. But if it doesn’t work on first explanation don’t give up. If x is even divide by 2 If x is odd multiple 3 add 1 gives you even so divide by 2 still even divide by 2 till equals 1 The conjecture shows a relationship in factors but does not show those factors. Let me know what you think. I am problem wrong in the understanding of this problem. But that is ok. It just leads to more discussion.

I am also an amateur mathematician. I agree your last example equals. But with all these posts the thread is out of control. I have read 3pages but all these changes confuse the reader. I know it is for your benefit to improve your proof, but I can’t understand it. I know it is bad practice to work with only limited examples, but could you show from the beginning an example? Here is an excerpt from the book Primes and Programing “The fundamental theorem 1.1 has a simple look about it, and indeed it is nice to know that every number greater than 1 is prime..or factors into primes. But actually finding the factorization…(or primality) can be very hard.” I request a through example because I don’t know what is the given and what we are solving. I think an example will show what you are describing and what variables we know. Can you write in 2 sentences what is the goal of the Proof. (A short abstract) Again I am also an amateur mathematician but for your proof to work everyone must understand what you are attempting to show. I like to work with examples when trying to understand series.

Are you saying that to add 1 to a number divisible by 2 will make it odd? And if you can determine if this odd number is a semi-Prime the Prime factors must be less than 2 to the power of the semi-Prime? My question is what way did you use the conjecture to solve the problem of factoring? This will increase in difficulty with larger numbers. Lemme 2, lines 48 & 49 lost me.

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.

Congrats on the minor discovery. Any discovery however little, is significant. But the next thing I would do is see where the discovery applies. It may seem like a simple polygon, but it is also a geometric construction. (What I am calling is using drawing and geometry to form shapes with measurable properties. Like dividing a line in half with 2 arcs of a compass.) I know it seems like the use for drawing this way is minor. However, what if this drawing method was applied to computer graphics. Imagine drawing a polygon mesh in 3DS Max and have that mesh be measurable by your polygons. It is difficult to come up with a major application, because this isn’t my design, but if you expand it to polygons of more sides with the inside of those polygons measured; you’d have a brick to build a polygon block and measure it at the same time. Now that isn’t a little discovery.

I hope this is the right place to post this, but where is imatfaal? I always enjoyed her comments. I wanted to get her opinion. I don't know her. She could be anyone, as this is an Internet forum. But I was wondering why she hasn't returned.

I was wondering this the other day. How many infinity symbols are there? There is the car Infinity whose symbol I'm not sure what infinity it is. I wonder if infinity has the most symbols representing it than any other math symbol. For example this is infinity:

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.

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

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.

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.

I don’t follow your method, but what I think you are trying to do is have an equation explain an unknown, irregular graph or shape. That would be gold dust that would allow any graph to be easily described. I think of it as 3D Studio Max. You are drawing a shape that is irregular from pieces of “molded” shapes. The drawing file in Max must be mathematically described, but is not an elegant equation describing the contours. But remember when you draw you draw starting with the basic shapes. As in your transformation from one graph to the other, it appears you have no reference graph. Perhaps you should try a known, defined graph and try to modify it to from the “transformed graph.” Remember in subjects like electricity, much descriptions are transformations of the Sine curve. I am not a math expert, obviously. However, I now you must simplify this problem. I don’t understand the transformation you did in your drawing. However, do not let that stop you, because how you explained it may be entirely different that the method I understood. But as I have personally learned, explaining the math problem is as important as finding the answer.

Ok I am going to suggest something that may be wrong, but it will show my thinking. If you take the Pi Angles here and multiply “every other” circular value by a number less than the next angle divided by the “every other” angle, you would get an ellipse on the graph. This would give an ellipse whose equation is represented in a different way than x^2/a^2 + y^2/b^2. You can build any circular function. I believe they are called conic sections. Does anyone have any thoughts on this? There are many series that could be used and in various patterns. Suppose the circle that the values were rolled out to the right, was a logarithmic spiral instead of an involute. Combine that with different combinations and different series and you could draw anything. If I am wrong or my explanation is confusing (which it is), I will clarify, but I believe anyone familiar with the coordinate plane will understand this. At this time, I can’t describe it well. But the idea is simple. Putting the math steps in words is difficult.

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.

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.

Good job. Keep up the great work. I don’t think it is a thousand years old. You have just used an angle and may have unknowingly drawn an involute which is a type of logarithmic spiral. (In case you don’t know I love logarithmic spirals.) When you unroll each outline to the right; Can someone in this forum tell me if it is a linear representation of the involute? I have seen something graphical, similar to this unroll in a math reference. I can’t remember where, but I believe it was for gears. I don’t have any pictures of a logarithmic spiral to share yet. I want to be sure it relates to your post. But even if this work is rediscovered, it doesn’t mean you can’t relate it to something new. What I think you should try is to “put space between your Pi angles.” What I mean by this is that having a “series” between where 1/3 r and 2/3 r and r would change the shape of the involute to a special logarithmic spiral. I know it sounds like I’m talking babble, but I am not. If you are confused on what I am trying to say, let me know and I will try and describe my idea better. Simple put I would shift the new larger angles a distance (determined by a series) across the x-axis from the original triangle that was at the origin. This way you can craft series and describe them in a logarithmic spiral. I will post a picture of an ellipse determined by angles. It is not a logarithmic spiral, but it will demonstrate using angles to determine geometry. I will try to work on drawing a graphic representation of the logarithmic spiral I describe here. But this will work till then.