But it's not complete yet. Your last case, a+b < 2c, a < c < b (the one you left for me to do), is the only one that isn't obvious.

If I substitute a = 1, b = 4, and c = 3, which satisfy those two inequalities, in

(a+b)[(a+b)^2 + 3 (a-b)^2] :: (2c)^3,

I get 5(5^2 +3·3^2) = 260 for the left member and 6^3 = 216 for the right one. The expression on the left has a greater numerical value than the one on the right.

If I substitute a = 1, b = 8, and c = 7, which also satisfy both inequalities, I get 9(9^2 + 3·7^2) = 2052 for the left member and 14^3 = 2744 for the right. Now the expression on the right is numerically greater.

(Of course it's obvious that neither triad of numbers satisfies the equation a^3 + b^3 = 2c^3. There are no solutions, as *I* have already proved.)

But what this numerical example shows is that those inequalities are *not* inconsistent with the equation. In other words, if there *were* any integer solutions a, b, c (though we know that there aren't), they would satisfy those inequalities.

So I'll have to ask you to complete the proof for me. I'm not up to it.

But it's not complete yet. Your last case, a+b < 2c, a < c < b (the one you left for me to do), is the only one that isn't obvious. ...> >

Yes, but it is junior high school stuff that you *mastered*. Remember your arrogant response to me? This is the return ticket. Do it or admit your incompetence.

<If I substitute a = 1, b = 4, and c = 3, which satisfy those two inequalities, in

(a+b)[(a+b)^2 + 3 (a-b)^2] :: (2c)^3,

I get 5(5^2 +3·3^2) = 260 for the left member and 6^3 = 216 for the right one. The expression on the left has a greater numerical value than the one on the right.

If I substitute a = 1, b = 8, and c = 7, which also satisfy both inequalities, I get 9(9^2 + 3·7^2) = 2052 for the left member and 14^3 = 2744 for the right. Now the expression on the right is numerically greater.>

Oh my, I claimed that equality does not hold. Your examples (like those *counterexamples* you gave before in another situation) show that you do not understand plain English or the problem. You always get lost in non-relevant and selfconstructed details losing sight of the problem and what was said.

<So I'll have to ask you to complete the proof for me. I'm not up to it.>

Hey, are you admitting that you are not able to do junior high calculations? Take your time.

<Wouldn't it have saved both of us some time if you had explained it fully in the first place?>

I did. Ok, when I gave the hint I assumed that most mathematically inclined people would see it. I was overestimating your abilities. After that we have your permanent and seemingly deliberate "misunderstandings" of basics.

You've spent all that time proving how smart you are, and you're not going to finish the job?

I guess the proposition you're trying to prove isn't true.>

I posted a *complete proof* that there are no nontrivial integer solutions of a^3 + b^3 = 2c^3.

Oh my, <princeton joker>. I never made that proposition. However, I proved what a hot air balloon you are. Anyway, we should put this into the "memorable quotes":

<<al wazir: ...
You don't belong on this page. You're not smart enough.>

Let a ≠ b,

a^3 + b^3 = 2*c^3 <=> (is equivalent to)

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

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

(a+b) =(2*c)*(p^2 + 3*q^2)

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

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

thus be have:
1.) a+b = 2*c*(p^2 + 3*q^2)
2.) (a+b)*p + 3*q*(a-b) = 2*c
3.) (a-b)*p = q*(a+b)

substituting 2.) in 1.) gives

(a+b)*(1-p*(p^2 + 3*q^2) =
(a-b)*(3*q*(p^2 + 3*q^2)

For p=1 and q=0 both sides are 0.
and 2.) gives a+b = 2*c

For all q ≠ 0 we have p≠ 0 because of 3.) and the assumptions about a and b.

Thus a-b=(q/p)*(a+b)

Substituting this in 2.) gives with 1.)
(a+b)((p^2+3*q^2)/p) = 2*c*((p^2+3*q^2)^2)/p = 2*c

or

a+b<2c and a+b=2*c*(p^2 + 3*q^2)

with

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

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

implies

p^2+3*q^2<(1/2)

Your "proofs" are still incomplete. As you've defined p and q, you haven't shown that they are integers.

Also,

a+b<2c and a+b=2*c*(p^2 + 3*q^2) imply p^2+3*q^2<1, not p^2+3*q^2<1/2. (I'm sure that's what you meant.) But there is a contradiction only if p and q are integers.

Finally, you may not have noticed that if we set

2u = a + b and 2v = a - b,

which I wrote as Eq. (3) in Louis Stumpers (kibitz #7988), then your equation

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

becomes

u(u^2 + 3v^2) = c^3,

which was my Eq. (4).

In that post I went on to prove that there are no nontrivial integer solutions to a^3 + b^3 = 2c^3, *without* making the assumptions a + b < 2c or a < c < b.

That is the very same elementary blunder that I made, for which you rightly criticized me.

Your "proofs" are still incomplete. As you've defined p and q, you haven't shown that they are integers. ...>

If p and q are integers p^2+3q^2 is an integer >1 if q not 0. Which means a+b>2c which is impossible (case 1)

You need to take some time off.

Suppose p = 3/5 and q = 1/5. ...>

I said p and q can be rationals.

You really are spouting garbage.>

<al wazir: ...

Your "proofs" are still incomplete. As you've defined p and q, you haven't shown that they are integers. ...>

there is only one case they are integers p=1 and q=0 (by the integers form a subset of rationals, don't you know?) Anyway, the garbage come from you <Your "proofs" are still incomplete. As you've defined p and q, you haven't shown that they are integers>