20191023, 00:18  #12  
May 2019
113 Posts 
Quote:
The expressions you were talking about were not polynomials in d and e, but you defined d and e to be integers, not polynomial variables, hence you cannot just compare coefficients. Not true in general does not mean there does not exist some h=i such that the expression somewhat holds by fluke, especially in the case where n is odd when the terms have alternating sign and they somehow cancel each other. Just one counterexample and the conjecture is false. Last fiddled with by 2M215856352p1 on 20191023 at 00:22 

20191023, 02:14  #13  
Undefined
"The unspeakable one"
Jun 2006
My evil lair
5,879 Posts 
Quote:


20191023, 11:35  #14  
Feb 2017
Nowhere
3·19·67 Posts 
Quote:


20191023, 13:06  #15 
Feb 2019
1011011_{2} Posts 
I don't understand what you mean when you talk about d and e not being polynomial variables and therefore I cannot compare coefficients. Comparison of coefficients is justified in order to establish the equality of LHS and RHS of the equation since the expressions highlighted in red are the same for RHS and LHS. You also talk about counterexamples. Counterexamples of what? Whatever counterexamples you are talking about, produce them.

20191023, 13:08  #16 
Romulan Interpreter
Jun 2011
Thailand
2^{2}×7×11×29 Posts 

20191023, 14:29  #17  
Nov 2008
2·3^{3}·43 Posts 
Quote:
This is true for the specific values d = 2 and e = 3, for example. This does not allow us to conclude that 3 = 6 and 4 = 2. (Also from what I can tell the "proof" never used the fact that A, B and C are all integers?) Last fiddled with by 10metreh on 20191023 at 14:32 

20191024, 06:32  #18  
May 2019
113 Posts 
Quote:
A^{x} + B^{y} = (A^{x/3})^{3} + (B^{y/3})^{3} = [(A^{x/3}) + (B^{y/3})][A^{x/3})^{2}  (A^{x/3})(B^{y/3}) + (B^{y/3})^{2}] = M^{3} It is defined that M = d  e, A^{x/3} = hd, B^{y/3} = ie, where d and e are positive integers and h and i are positive real numbers, if I have not mistaken. By division by M from both sides, we end up getting [(hd + ie)/(d  e)]h^{2} d^{2}  [(hd + ie)/(d  e)]hi de + [(hd + ie)/(d  e)]i^{2} e^{2} = d^{2}  2de + e^{2} Comparing coefficients of d^{2}: [(hd + ie)/(d  e)]h^{2} = 1 Comparing coefficients of de: [(hd + ie)/(d  e)]hi = 2 Comparing coefficients of e^{2}: [(hd + ie)/(d  e)]i^{2} = 1 Hence, indeed there is a contradiction by comparing coefficients. However, weird examples like 2^2+11^2=5^3 can appear because it does not need to be the case where each term is equal in value for a specific value of d and e. A counterexample to Beal’s conjecture is very hard to find. Till now, we still have not found any yet. Please notify me if I made any errors. Awojobi's variable names spell hide, he seems to be hiding something, isn't he? (joking) Last fiddled with by 2M215856352p1 on 20191024 at 07:29 

20191024, 08:07  #19  
Jun 2003
7×683 Posts 
Quote:
However, while the proof itself say that A,B,M, x,y,d,e are integers and h&i are rationals, none of the algebraic manipulations require it to be so (i.e. uses these properties). So we can pick arbitrary values for these  specifically, we can set M= (A^x+B^y)^(1/3) and observe the behavior of the coefficients. OP will handwave these away, but that's SOP for him. Code:
A=5 B=7 x=3 y=3 M=(A^x+B^y)^(1/3) /// An M that works e=1; /// arbitrary d=e+M h=A/d i=B/e c1=(A+B)/M*h^2 ///x/3 = y/3 = 1 c2=(A+B)/M*h*i c3=(A+B)/M*i^2 d^22*d*e+e^2 c1*d^2+c2*d*e+c3*e^2 /// Look at that! Eventhough coefficients are different, end value is same! Last fiddled with by axn on 20191024 at 08:14 Reason: Example 

20191024, 12:06  #20 
Feb 2019
7×13 Posts 
So 2M215856352p1 we are in agreement. The example you gave doesn't count be cause 2 of the powers are 2 and Beal's conjecture clearly states that the powers should be greater than 2. The other poster coming up with 3d+4e = 6d+2e to try and debunk my rationale for equating coefficients should note that the equation simplifies to 3d = 2e. There are no coefficients to compare now. Also, it is clearly stated in the Beal equation that A, B and C are all integers.

20191024, 13:13  #21 
Feb 2017
Nowhere
3·19·67 Posts 

20191024, 13:17  #22  
May 2019
113 Posts 
Quote:
On my part, it was a completely failed attempt to come up with a counterexample. I actually disagree with you. I still have one question to ask: why is it necessary that x, y, z, M are positive integers? It is stated in the conjecture but you have yet to use those properties in your proof. axn found a counterexample in your logic of the proof but not the conjecture. Hence I need you to explain why the logic in your proof works when x, y, z, M are positive integers, x,y,z>2 without stating that it is in the conjecture. Otherwise, that would be a circular argument. 

Thread Tools  
Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
A Flawed Proof of Fermat's Last Theorem  paulunderwood  Miscellaneous Math  16  20191021 23:40 
Proof of Fermat's last theorem h^n+k^n=N^n with N even  Alberico Lepore  Alberico Lepore  7  20191020 21:06 
The Beal Conjecture Proof  Arxenar  Miscellaneous Math  1  20130907 09:59 
abc conjecture and Fermat's Last Theorem  jasong  jasong  3  20121024 08:45 
Proof of Fermat's Last Theorem  McPogor  Miscellaneous Math  18  20071019 11:40 