PROOF OF BEAL'S CONJECTURE

[dodomaze]

[tex]\cancel{C^z - B^y = \left( C^{z/n} - B^{y/n} \right)^n}[/tex]
Wow, big mistake, my apologies. I tried to copy (badly) the typesetting from a previous post, and then ran blindly with the mistake. Totally my bad.

Solve the equation (a^0.5 - b^0.5) (c^0.5 + d^0.5) = 3^2 where a,b,c,d are integers. In order to solve this equation, is it not clear that only integer values of a^0.5, b^0.5, c^0.5 ,d^0.5 should be considered and not irrational numbers?

I don't see why. Taking a=c and b=d produce an infinite number of solutions as long as [tex]a-b=3^2[/tex] (maybe you meant 3 as the right-hand side, but we can go with 9 as well). So a=c=11 and b=d=2 works, but a = c = [tex]11-\sqrt 2[/tex] and b = d = [tex]2 + \sqrt 2[/tex] would do as well. And I can't discard a priori solutions where a [tex]\ne[/tex] c and b [tex]\ne[/tex] d.

Back to what I wanted to say, before I blundered big time: you claim that
[tex]C^z - B^y = \left( C^{z/n} - B^{y/n} \right) \cdot \left( (C^{z/n})^{n-1} + (C^{z/n})^{n-2} B^{y/n} + (C^{z/n})^{n-3} (B^{y/n})^2 + \ldots + (C^{z/n})^2 (B^{y/n})^{n-3} + C^{z/n} (B^{y/n})^{n-2} + (B^{y/n})^{n-1} \right)[/tex]

(if I'm not mistaken this time) cannot be equal to [tex]A^x[/tex], because the right-hand side cannot be an integer. But the left-hand side, [tex]C^z - B^y[/tex], is an integer!

You are able to see an algebraic cancellation that justifies the equality to the integer [tex]C^z - B^y[/tex], but not seeing one for [tex]A^x[/tex] is no justification that the expression cannot produce such an integer. Let alone an integer, especially when you started from an integer when constructing this expression. It gives the impression that the definition of equality is bent, to mean one thing or another depending to what is convenient.

[Awojobi]

Okay, you showed that there are irrational number solutions which I assumed were non existent. Basically, this equation I formulated has irrational number solutions and integer solutions, just like my original equation i.e. the equation you wrote at the end of your last post. Let's consider the irrational number solutions in the equation I formulated and even the main equation you wrote in your last post. If we work out the numerical value in each bracket, we get rational numbers to whatever degree of accuracy we wish. So, the product of the 2 rational numbers gives an answer as close to 9 as we desire but never actually being equal to 9, considering the equation I formulated. So, to answer the last part of your question in your last post, yes equality can be looked at as exact and tending towards exact, depending on whether expansion of brackets is carried out or products of the brackets are carried out. The reasoning behind my proof, which I probably should have written down in the proof, is that the product of the 2 brackets should be seen as a first step in trying to do a factor decomposition. Factor decomposition consists of integers and not irrational numbers. The proof is trying to show that the product of the 2 brackets is an integer, A^x, which is a product of its factors, so what better way to do this except trying to breakdown the product of the 2 brackets into its integer factors. This is why irrational numbers need ever be considered in the proof of Beal's conjecture but only integers.

[dodomaze]

As an aside, I suspect you may have been using a computer too much lately. Keep in mind that computers typically do calculations only with rationals (namely, of the form [tex]m \cdot 2^n[/tex], as per the IEEE 754 standard). This tend to give the impression that these are the only numbers in existence.

Computation belongs to applied mathematics. Pure mathematics is perfectly happy doing algebra with irrationals. As much as you'd wish to decompose your integer into smaller integer factors (you could at least have "x" of them, in the worst case where A happens to be a prime), ignoring irrationals is not an strategy; I already showed examples of addition and multiplication of irrationals that produce integers, and you keep discarding them either by calling them "trivial" or by insisting on a non-exact equality.

The problem with equality is a fatal flaw in your proof. The ability to say that either A = B or A [tex]\ne[/tex] B, and that only one of these is true (either one or the other, but not both) is a cornerstone of mathematical logic, and invalidating it breaks any attempt at proving anything. The entire building of mathematics would collapse if an equivalence relation like "=" is broken.

[Awojobi]

Concerning your last statement, let's therefore discard what I was saying about equality but concentrate on the reason I gave about why the terms in each of the 2 brackets MUST be integers so that each bracket never works out as irrational. I will repeat it here again and as I said earlier, I should have written it in my proof for clarity. So, each bracket MUST work out as integers so that their product can undergo factor decomposition because factor decomposition is the first step in the proof. Irrational numbers therefore need to be ignored because they clearly can't meet the criteria for factor decomposition which deals only with integers.

[dodomaze]

need to be ignored because they clearly can't meet the criteria for factor decomposition which deals only with integers

That is not clear at all. Irrationals can be added or multiplied together to produce integers. Insistence on the contrary comes only because of the broken equality.

- - -

P.S.: If you were referring to the unique factorization of integers into primes, then you are heading into even deeper trouble.

[tex]C^z - B^y[/tex] is the difference of two coprime numbers. The prime factorization of that difference is nearly unpredictable (with respect to the prime factorizations of the two terms [tex]C^z[/tex] and [tex]B^y[/tex]); the only way of finding it is just to "undo" the original factorizations, subtract the two numbers, and create the factorization of the result separatedly, with no relation to the factorizations of the individual terms. Obtaining the factors of the result directly from the factors of the two terms is an even larger problem than Beal's conjecture, and I don't see any attempt at it in your proof.

Claiming that your workout shows these factors to be inexistent, would be like claiming that 15 cannot be expressed as the product of two primes, because you can write 15 = 64 - 49 = 8^2 - 7^2 = (8 + 7)(8 - 7), and the latter is always 1. This simply shows that the factors of 15 cannot be found this way, not that they cannot be found at all.

[dodomaze]

A brief detour: you are probably familiar with quadratic equations, like [tex]x^2 - 5x + 6 = 0[/tex]. To make matters simple, I will consider only equations of the form
[tex]x^2 + bx + c = 0[/tex], where b,c are integers, and the coefficient of [tex]x^2[/tex] is always 1.

These equations have always two solutions, two "roots" - two numbers that make this polynomial equal to zero. These roots can be integers, rationals, irrationals or even complex numbers. (They could be two separate numbers, or the same number twice, but that is irrelevant to my argument.)

Call these two solutions [tex]x_1[/tex] and [tex]x_2[/tex]. Since any of these values would make the polynomial equal to zero, an alternative expression for this polynomial is
[tex]x^2 + bx + c = (x - x_1)(x - x_2)[/tex], where it is more clear how the expression is zero when either [tex]x=x_1[/tex] or [tex]x=x_2[/tex].

Expansion of parentheses on the right-hand side leads to
[tex]x^2 + bx + c = x^2 - (x_1 + x_2)x + x_1 x_2[/tex]

which illustrates how you can take [tex]b = -(x_1 + x_2)[/tex] and [tex]c = x_1 x_2[/tex] and obtain the same polynomial. But b,c were integers! And [tex]x_1, x_2[/tex] could be irrational, or complex!

These are called "Vieta's formulas", and have been in use since the late Renaissance. It may be amusing to watch Veritasium's video in YouTube titled "How Imaginary Numbers Were Invented", or "a duel between mathematicians" (it concentrates on imaginaries rather than irrationals, but the arguments are similar to the ones we are having here).

[Awojobi]

Let's look at a simpler equation in order for me to make my point. a^0.5 + b^0.5 = c, where a,b,c are positive integers, a and b don't share any common factors except 1 i.e. they are co-prime. This immediately means that we do not need to consider the irrational values of a^0.5 and b^0.5 but only integer values if we want to look for solutions to the equation for a given value of c. This is exactly the reasoning I have used in my proof.

[dodomaze]

You have constructed a particular example that, indeed, has no solutions other than a,b being perfect squares. But that specific case is not directly related to your proof. Plus, the examples you are trying to avoid with your conditions are not the only ones.

You could even let a,b not be integers at all. Then choose an arbitrary real number [tex]x[/tex], and use [tex]a = x^2[/tex] and [tex]b = (c - x)^2[/tex], and the sum of [tex]\sqrt a + \sqrt b[/tex] would be equal to [tex]c[/tex].

As an example, for [tex]c = 5[/tex], we can arbitrarily choose [tex]x = 3^{1/7}[/tex], and then use
[tex]a = 3^{2/7}[/tex] and [tex]b = (5 - 3^{1/7})^2 = 25 - 10 \cdot 3^{1/7} + 3^{2/7}[/tex],

and we would have [tex]\sqrt{3^{2/7}} + \sqrt{25 - 2 \cdot 3^{1/7} + 3^{2/7}} = 5[/tex].

See the result in Wolfram Alpha

No reason to discard irrationals you don't have any details about, simply because they are irrationals.

[Awojobi]

It seems your last equation has an error i.e. 2 should be 10. In any case, I can't see the significance in what you are trying to show. I have shown an example that relates to my proof and you discard it.

[dodomaze]

Yes, thanks for noting the mistake.

Maybe it would be a good time to write some corrections into your text, and make something more complete. You have been mentioning factorization, and it's still not clear what you mean exactly by that. You claim the example on your previous post is related to the proof, but it is not explained how. It may be the time to gather all the notes you have been adding here into a coherent whole.

[Awojobi]

Maybe I should come up with the concept of irrational factors. For instance, 5^0.5 is an irrational factor of 5. 2^1/3 is an irrational factor of 2. So, if both brackets of the factorised form of the Beal equation are irrational, they can be considered as irrational factors. We are not concerned with the irrational factors when attempting to prove Beal's conjecture. We are interested in the integer factors, hence only integer factors should be considered. Therefore, every term in each of the 2 brackets must be integers.