April 19th, 2019, 05:31 PM  #1 
Member Joined: Mar 2019 From: california Posts: 40 Thanks: 0  Fermat Last Theorem Diophantine Analysis
Analysis using Diopphantine Equations Diophantine Equations are used to remove as many variables as possible and write the remaining unknowns in terms of the other unknowns. By analyzing the remaining terms as whole numbers, we can decided whether there are infinite number of solutions or zero solutions. Because we have the given equation X^n + Y^n = Z^n. From this equation we can derive to associated equations 1) X + Y = c/d * Z where c/d > 1. c,d are rational also c/d < 2^1/2 2) X^2 + Y^2 = e/f * Z^2 where e/f > 1 and c/d > e/f e,f are rational from 1) we get d*X + d*Y = C*Z] d*X = c*Z  d*Y or finally X = c*Z/d  Y If we substitute value for X into 2) we get ( c^2/d^2 * Z^2  2c/d * Z * Y + Y^2) + Y^2 = e/f * Z rearranging c^2/d^2 * Z^2  2c/d * Z * Y  e/f*Z = 2Y^2 or Z *(c^2/d^2 * Z  2c/d * Y  e/f) = 2Y^2 Now let Z = h *Y where 2 > h > 1 and substitute into equation h * Y *((c^2/d^2 * h *Y)  2c/d * Y  e/f) = 2Y^2 factor Y out of both sides of equation and you get h *((c^2/d^2 * h *Y)  2c/d * Y  e/f) = 2Y Now move 2Y to left hand side and extend h across members of brackets c^2/d^2 * h^2 *Y  2c/d * h * Y + 2Y  e/f * h = 0 move e/f * h to right hand side. Now left hand side has the Y come out of all terms on left hand side (c^2/d^2 * h^2  2c/d * h + 2) * Y = e/f * h or finally Y = (e/f * h)/(c^2/d^2 * h^2  2 * c/d * h + 2) We do not know what the exact value of y is, but we can determine the range when Y is close to Z, e/f approximates 1, h approximates 1, c^2/d^2 approximates 1, but is always less than 1 In that case Y is approximately (1 * 1)/((1 * 1^2)  2 * 1 * 1 + 2) = 1/1 = 1, but less than 1 When Y is close to X, e/f approximates 1.3, h approximates 2, c^2/d^2 approximates 1.4 Then Y is approximately (1.3 * 2)/(1.4^2 * 2^2  2 * 1.4 * 2 + 2)= 2.6/7.84  5.6 + 2 or Y is approximately 2.6/4.24 or approximately 0.6 These approximations could be made more defined, but the real issue has already been decided. The range of values Y takes on is between 0.6 and 1. There is no way Y could be equal to 2, even if the numbers are refined. It should be noted that Y is the larger of the supposed values X,Y. This only leaves an integer value of less than 0 for X, which is not allowed. This completes the proof for Fermat's Last Theorem using Diophantine Equations. 
April 19th, 2019, 10:55 PM  #2 
Senior Member Joined: May 2013 Posts: 118 Thanks: 10 
in this equation you give: ( c^2/d^2 * Z^2  2c/d * Z * Y + Y^2) + Y^2 = e/f * Z you should write Z^2 instead of Z in right hand side Check if this blows everything up or not 
April 20th, 2019, 11:33 AM  #3  
Senior Member Joined: Aug 2012 Posts: 2,306 Thanks: 706  Quote:
Proof busted. Your CLAIM that n > 2 (which you didn't bother to specify) must lead to $\frac{e}{f} > 1$ requires assuming FLT is true in the first place. This is the exact same error as in your first false proof. In that proof you CLAIM that alpha > 0 for n > 2 but it's perfectly clear that this assertion depends on the truth of FLT. Indeed, my earlier counterexample works here as well. If $x^{10} + y^{10} = z^{10}$ then you have a Pythagorean triple $(x^5)^2 + (y^5)^2 = (z^5)^2$ and $\frac{e}{f} = 1$. The only way you can prove that can't happen is by assuming FLT in the first place. By the way I would appreciate your reply in your previous thread with the infinite descent of triangles. I pointed out that you do indeed have all those triangles, but they all have irrational sides and you haven't proved anything. I await your comment. Last edited by Maschke; April 20th, 2019 at 11:58 AM.  
April 24th, 2019, 02:19 PM  #4 
Member Joined: Mar 2019 From: california Posts: 40 Thanks: 0 
I did reply to your comment about the infinite descent of triangles. Let me know if you now have it. There is no assumption that FLT is true for either equation 1 or 2 Equation 1 is derived from the fact that only the initial equation is true.There is the assumption that the values of X,Y,Z are rational. It really doesn't matter whether the numbers are rational or not. . This is purely a consideration of the number line and where this numbers are on the number line in relation to each other. That means that Z is larger than either X and Y. If Y is the larger of the two, than Z must be less than 2Y. ( I made a notation error and said c/d <2^1/2.The reasoning still holds. The range for Y is still is between 0.5 to 0.6 approximately.) Equation 2 is derived from the fact that the squares of the numbers X,Y,Z have a a similar relationship to each other, the same as equation 1. X^2 + Y^2 = e/f *Z^2 If X was as big as Y, then Y^2 +Y^2 = Z^2. 2Y^2 =Z^2 or Z =2^1/2 Y So Z = e/f Y means that the range of Z is between 1 to 1.4 (2^1/2). As a side note, when n is 2, then e/f is 1 as you say. That's why Fermat specified that n be greater than 2. 
April 24th, 2019, 08:50 PM  #5  
Senior Member Joined: Aug 2012 Posts: 2,306 Thanks: 706  Quote:
Easy proof of Fermat's last Theorem In that thread you start with an integer right triangle and then keep dropping a new triangle underneath it whose hypotenuse is the lower leg of the previous triangle. After doing this a number of times, you triumphantly say, "And see, some triangle has an irrational side." If I'm understanding you correctly. And then you say that proves FLT. There are two problems. First, every side of every one of your triangles is irrational, except in exceptional cases where the sides of the first triangle are all squares. Even so you will soon get to the case where all your sides of all your triangles are irrational. So far from proving that some side will eventually be irrational, all the sides of all the triangles are irrational. But regardless, I don't see how this proves FLT. You haven't explained that. I do await that explanation. Quote:
Quote:
Quote:
Quote:
Quote:
e/f = 1 for n = 10 as well. Your false claim that e/f > 1 is the same as your first thread when you claimed alpha > 0. The exact same example of n = 10 shows that alpha = 0 whenever n is even.  
April 25th, 2019, 05:04 PM  #6  
Member Joined: Mar 2019 From: california Posts: 40 Thanks: 0  Quote:
Y = (e/f * h)/(c^2/d^2 * h^2  2 * c/d * h + 2) is the final result To determine the range of Y we have to determine what values Y takes on when Y is close to Z, and when Y is close to X When Y is close to Z e/f close to 1 h is close to 1 c^2/d^2 close to 4 c/d is close to 2 Y = 1*1/4*12*2+2 = 1/2 Y = (e/f * h)/(c^2/d^2 * h^2  2 * c/d * h + 2) Now when X is close to Y e/f is close to 1.4 h is close to 1.4 c^2/d^2 close to 4 c/d close to 2 Y = 1.4 * 1.4 / 4*2  2*2*1.4 +2 or Y = 2/ 85.6 + 2 or Y = 2/4.4 = 0.45 So the range of Y is 0.45 < Y < 0.5 Now, since Y must be less than 1, FLT is proved.  
April 25th, 2019, 05:16 PM  #7 
Senior Member Joined: Aug 2012 Posts: 2,306 Thanks: 706  I have no idea what you're doing. Is e/f still claimed to be greater than 1? How does any of this prove FLT? Where did you use the assumption that n > 2? It's the classic hallmark of a false proof of FLT that the claimed proof does not distinguish the case n = 2. What are X, Y, and Z? I had assumed they must be integers but at some point you said they're rationals. You started in the middle of your argument here. I don't know what anything is supposed to be. Earlier you said e/f > 1 but now you've abandoned that idea without admitting you did or explaining that you didn't. Last edited by Maschke; April 25th, 2019 at 05:33 PM. 
April 26th, 2019, 12:05 PM  #8 
Member Joined: Mar 2019 From: california Posts: 40 Thanks: 0  Here is the diagram that Fermat probably used to demonstrate the relationships between X<Y>Z<X^2 +Y^2, Z^2, Z^n, X^n +Y^n. When n > 2, Z^2 < X^2 + Y^2 every time, it doesn't matter which n you talk about. When n = 3 and you use the number 3,4,5 then e/f is greater than 1 and we can calculate the value of e/f exactly. If we use different numbers, then the value of e/f will change but always be in the range. Last edited by skipjack; April 26th, 2019 at 12:52 PM. 
April 26th, 2019, 12:50 PM  #9 
Global Moderator Joined: Dec 2006 Posts: 20,623 Thanks: 2076 
The diagram would need to be uploaded to this site. Also, you seem to be ignoring most of the questions that you have been asked, and unable to express your reasoning clearly.

April 26th, 2019, 03:49 PM  #10 
Senior Member Joined: Aug 2012 Posts: 2,306 Thanks: 706  No this is not true. If there were a Fermat triple for n = 10 you would have a right triangle with sides $x^5$, $y^5$, and $z^5$ and e/f would be exactly 1. You simply refuse to engage with this point but it breaks your proof.


Tags 
analysis, diophantine, diopphantine, fermat, theorem 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Fermat's last theorem  Lourie  Number Theory  3  April 1st, 2017 12:37 AM 
About Fermat's Little Theorem  McPogor  Number Theory  5  December 7th, 2013 07:28 PM 
Fermat's Last Theorem  mathbalarka  Number Theory  2  April 3rd, 2012 11:03 AM 
More Fermat's Last Theorem.  theomoaner  Number Theory  29  November 26th, 2011 10:23 PM 
Fermat's last theorem.  SnakeO  Number Theory  10  September 25th, 2007 04:23 PM 