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Old 2019-10-23, 00:18   #12
2M215856352p1
 
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Originally Posted by Awojobi View Post
It is not an assumption. It is justified because the 2 sides of the equation have the same corresponding expressions highlighted in red. It therefore means equating corresponding coefficients is justified. Of course when this is done, contradictions galore begin to arise which shows that the original assumption of an equation is contradicted, given the conditions stated in the proof. Herein lies the proof.
Ultimately, this is a wrong justification. This is a rather glaring mistake if you are observant enough.

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 2019-10-23 at 00:22
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Old 2019-10-23, 02:14   #13
retina
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My proof ... would not even be looked at because I am not a respected professional mathematician.
You have no proof, therefore you get no looks.
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The purpose of me posting here is for a good critique ...
If you were more willing to accept critique instead of shouting people down then people might be more willing to post some.
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Old 2019-10-23, 11:35   #14
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Originally Posted by 2M215856352p1 View Post
Quote:
Originally Posted by Awojobi View Post
It is not an assumption. It is justified because the 2 sides of the equation have the same corresponding expressions highlighted in red.
Ultimately, this is a wrong justification. This is a rather glaring mistake if you are observant enough.
Yes, indeed! Everyone knows that, in order for a proof by highlighting to be valid, highlights should be in yellow!
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Old 2019-10-23, 13:06   #15
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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.
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Old 2019-10-23, 13:08   #16
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Old 2019-10-23, 14:29   #17
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Originally Posted by Awojobi View Post
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.
Consider the equation 3d+4e = 6d+2e.
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?)

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Old 2019-10-24, 06:32   #18
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Originally Posted by Awojobi View Post
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.
Let n=3. It follows that,
Ax + By = (Ax/3)3 + (By/3)3 = [(Ax/3) + (By/3)][Ax/3)2 - (Ax/3)(By/3) + (By/3)2] = M3

It is defined that M = d - e, Ax/3 = hd, By/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)]h2 d2 - [(hd + ie)/(d - e)]hi de + [(hd + ie)/(d - e)]i2 e2
= d2 - 2de + e2

Comparing coefficients of d2: [(hd + ie)/(d - e)]h2 = 1
Comparing coefficients of de: -[(hd + ie)/(d - e)]hi = -2
Comparing coefficients of e2: [(hd + ie)/(d - e)]i2 = 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 2019-10-24 at 07:29
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Old 2019-10-24, 08:07   #19
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A counterexample to Beal’s conjecture is very hard to find.
Understatement of the century. If we ever do find a counterexample, the conjecture will be falsified. So asking for a counterexample is to falsify the conjecture itself and not just the proof.

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^2-2*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 2019-10-24 at 08:14 Reason: Example
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Old 2019-10-24, 12:06   #20
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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.
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Old 2019-10-24, 13:13   #21
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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.
Hypotheses in Beale's conjecture don't count if your argument doesn't use them.
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Old 2019-10-24, 13:17   #22
2M215856352p1
 
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Quote:
Originally Posted by Awojobi View Post
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.
Not exactly, your proof is incomplete and I can sense some ego in your dismissal of the poster who sent 3d+4e=6d+2e. You can compare coefficients in 3d=2e, it’s just meaningless, 3=0 and 0=2.

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.
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