Yes.>

What is wrong with that???

comparing the corresponding terms it gives you:

(a+b)^3 = (2*c)^3 and (a+b)*3*(a-b)^2=0

But you seem to have compared f(f^2 + 3g^2) and f'(f'^2 + 3g'^2) term by term and concluded that f = f' and g = g'.

I don't see how you conclude that.>

Let a = 20 and b = 18. Let f = a + b = 38 and g = a – b = 2.

Let a' = 24 and b' = 2. Let f' = a' + b' = 26 and g' = a' – b' = 22.

Then f(f^2 + 3g^2) = 55328 = f'(f'^2 + 3g'^2).

oh man, you must be under shock after your bombast was brought to nothing.

We are discussing a^3+b^3=2*c^3. And in this particular case you can proceed as I did.

Really, you do not <... need help from Mr. Euler and Mr. Binet after all.> Better ask <Niesjesram> or <Abdel>.

121^3 + 137^3 = 23^3 + 163^3 = 4342914.

Let f = 258, g = 16; f' = 186, g' = 140.

1. <jb> advises another frequent poster to "stick to recreational mathematics." Does this mean that <jb>. has discovered a branch of mathematics that is no fun? I'd like to what <johnl> thinks of that.

2. <johnl> uses a mathematical wonder "to evoke a sense of awe at a universe where ...". What universe is that? Is there a mapping between mathematics and the natural universe? Which way does it go, or is it both ways?

<al wazir: 251^3 + 333^3 = 17^3 + 375^3 = 52739288

In this case f(f^2 +3g^2) is divisible by 4, so there's your alternative factorization.>

are you just playing stupid?
If you have 2 factors X and Y with
X*Y = 4*a^3 + 4*b^3 = (a+b)((a+b)^2 + 3*(a-b)^2) you have especially X*Y = (a+b)((a+b)^2 + 3*(a-b)^2). Did you get that?

1. <jb> advises another frequent poster to "stick to recreational mathematics." Does this mean that <jb>. has discovered a branch of mathematics that is no fun? ...>

Recreational mathematics for me is what you find e.g. in Martin Gardner's books. Easy to digest on weekends with a glas of wine.

No. You're just repeating yourself.>

Isn't Stumpers a recreational page, or do you have a professional mission to enlighten and impress, even here?>

1. make it taught by an "authentic" blowhard.

You posted a "hint" (Louis Stumpers (kibitz #7989)), which I didn't understand. (It's nothing more than a restatement of the formula for the factorization of the sum of two cubes, which I learned in junior high school.)

I asked a question. Your answer didn't make it clearer. Maybe you really do have a simple proof in mind, but you explained it very poorly. I still don't understand.

<al wazir: ... (It's nothing more than a restatement of the formula for the factorization of the sum of two cubes, which I learned in junior high school.)>

Since you learned that so early in your life you may have missed that

(a+b)*((a+b)^2 + 3*(a-b)^2) = (2*c)^3

allows for the conclusion a+b = 2*c.

that is why I wrote

(2*c)^3 = (2*c)*((2*c)^2 + 3*0).

a+b > 2*c and a+b < 2*c are impossible which leaves a+b = 2*c which necessarily gives a=b. (Remember, we are considering positive integers before you bring up further mental crap)

Furthermore, you apparently do not have a clue how the Euler/Binet formulae were derived.

So, here is a free lesson for you:

Treating the equation

a^3 + b^3 = c^3 + d^3

they put A = a + b, B = a - b,
C = c + d, and D = c - d which gives
A*(A^2 + 3*B^2) = C*(C^2 + 3*D^2).

Now, we are considering the case c = d.

That simplifies the equation to

a^3 + b^3 = 2*c^3.

With the little trick multiplying the equation by 4 we have

4*a^3 + 4*b^3 = (2*c)^3 =
(2*c)((2*c)^2 + 3*0^2)

and without any substitution required we have

(a+b)*((a+b)^2 + 3*(a-b)^2) =
(2*c)*((2*c)^2 + 3*0^2)

from which we can conclude a=b (see above)

(something that slipped you attention completely.)

What is interesting is since you did not understand these simple manipulations you threw in examples like

<al wazir: And if you insist that a, b must both be odd, here is another example:

121^3 + 137^3 = 23^3 + 163^3 = 4342914.

Let f = 258, g = 16; f' = 186, g' = 140.

f(f^2 + 3g^2) = 17371656 = f'(f'^2 + 3g'^2), but f ≠ f' and g ≠ g'.>.

Clearly documenting that you are confused and trying to make a point that has nothing to do with the case in question.

You posted a "hint" (Louis Stumpers (kibitz #7989)), which I didn't understand. ...>

Well, that's a different statement from what you posted right after my hint (which was a proof for those in the know):

< al wazir: <john barleycorn>: Your last equation makes no sense. But I see the contradiction, and it's a lot shorter than my proof.>

<I see the contradiction> and then admitting

<al wazir: <john barleycorn: f=a+b must divide (2*c)^3 and if f is not a prime number it does not have to divide 2*c.> Yes, my argument was nonsense.

But you seem to have compared f(f^2 + 3g^2) and f'(f'^2 + 3g'^2) term by term and concluded that f = f' and g = g'.

I don't see how you conclude that.>

Yes, I made a mistake and admitted it. But you can't do that. The Euler-Binet formulae deal with Fibonacci numbers. They have zero connection with this problem. Maybe what you meant was the Stanford-Binet formulae?

You don't belong on this page. You're not smart enough.>

wow. that is all what you got in reply?

Maybe, some members here think differently. Don't know.

they put A = a + b, B = a - b,
C = c + d, and D = c - d which gives
A*(A^2 + 3*B^2) = C*(C^2 + 3*D^2).>

NONE of the articles on the Euler-Binet (properly, Binet's) formula or the Fibonacci numbers that I consulted make use of that equation. You may be confusing it with Euler's treatment of the n=3 case of Fermat's theorem, which I modeled my proof on, or (less plausibly) with Lagrange's identity, which is a special case of the *Cauchy-Binet* identity. (I looked it up in Weisstein.)

Thank you for providing a step by step exposition of your "proof." It is still equivalent -- as I said earlier -- to arguing that

f(f^2 + 3g^2) = f'(f'^2 + 3g'2)

implies f = f' and g = g'. I gave several counterexamples.

And alas, another thing we have in common.

NONE of the articles on the Euler-Binet (properly, Binet's) formula or the Fibonacci numbers that I consulted make use of that equation....>

Typical for your reading problems. I wrote Euler/Binet not Euler-Binet to avoid confusions. It did not work in your case, though. My apologies. ("the complete solution was found by Euler and simplified by Binet" as per 5th edition of "An Introduction to the Theory of Numbers",by Hardy and Wright (p. 199)

Please note, that I also wrote (and you may not have read it):

<Since you learned that so early in your life you may have missed that

(a+b)*((a+b)^2 + 3*(a-b)^2) = (2*c)^3

allows for the conclusion a+b = 2*c.

...

a+b > 2*c and a+b < 2*c are impossible which leaves a+b = 2*c which necessarily gives a=b. >

Since you skipped grade 2 let me show you (a,b,c denote positive integers):

1) a+b>2c =>

(a+b)((a+b)^2+3(a-b)^2)>(a+b)^3>(2c)^3
if a≠b. If a=b we have (a+b)^3>(2c)^3 right away.

Are you still with me <al wazir>? yes? good!

2. a+b =2c gives

(a+b)((a+b)^2+3(a-b)^2)=(2c)^3+(2c)3(a-b)^2=(2c)^- 3 ergo (2c)3(a-b)^2=0 and a=b

Piece of cake, isn't it?

3. a+b<2c
(without loss of generality we assume a<b. a=b is trivial and if b<a you know what to do <al wazir>, don't you?

a) a<c and b<c
then a^3+b^3<2b^3<2c^3