Is this a solution for the problem: $\ a^3 + b^3 = c^3\ $ has no nonzero integer solutions?

Question

Is this a solution for the problem: $\ a^3 + b^3 = c^3\ $ has no nonzero integer solutions?

Suppose $\ a^3 + b^3 = c^3,\ a,b,c \in \mathbb Z^*,\ $then:
$c^3 - b ^ 3 = (c - b)((c - b) ^ 2 + 3cb) = a ^ 3 \quad (1)$
We can assume that all variables are coprime, because $\ c - b\ $ divides $\ 3cb,\ a\ $ and $\ c - b\ $ doesnt divides $\ c,\ b,\ $ so
$c - b = 3 \quad (2),$
from $(1)\ $ and $\ (2)\ $ get $\ 3 (3 ^ 2 + 3 c(c - 3)) = 3^{3}x ^{3},\ c ^ 2 - 3c + 3 = 3x ^3$,
here we see $3$ divides $\ c,\ $and we know $3$ divides $a$, this conflict by assuming.

Edit:

As Nishant commented: "I don't see why $\ c−b\ $ divides $\ 3cb$..."
Divide both side of $(1)\ $ by $\ (c - b)\ $ get $\ 3cb = (c - b)^{2}(x^{3} - 1)$

Update:

If$~(c−b)~$ is a single prime or a product of distinct primes or $~(c−b)~\nmid~a~$ and $~(c−b)~$ isn't a cubic number, then $~(c−b)~$ contains factor$~m~$ of $~a,~$divide both side of $(1)\ $ by $\ (c - b):$

$(c - b) ^ 2 + 3cb = (c-b)^{2}x^3 \quad (2),$

from $~(2)~$ if $~m=3~$ or not, we can get $~c~$ or $~b~$ contains factor $~m,~$this conflict by assuming.

If $~(c−b)=1~$ then $~3c^2-3c+1=a^3,~$

from Wolframalpha get:

$$ c = \dfrac{3- \sqrt{3}\sqrt{4a^{3}-1}}{6} \\ c = \dfrac{\sqrt{3}\sqrt{4a^{3}-1}+3}{6} $$ There's no integer solution (As Steven Stadnicki commented,$~\sqrt{3}\sqrt{4a^{3}-1}~$ isn't an integer, lack of proof)(update: this solved by Jack D'Aurizio see: How to prove $~\sqrt{3}\sqrt{4a^{3}-1}~$ isn't an integer?).

(If$~(c−b)~$ is a cubic number, the problem left: $~(c - b) ^ 2 + 3cb=x^3~$has no nonzero integer solutions for $~c,~b$)

Answer 1

Even if $a,b,c$ are relatively prime, $c-b|a^3$ does not necessarily imply that $c-b |a$.

A simple counterexample is $c=13, b=5, a=2$.

Your conclusion (1) is incorrect.

Answer 2

The problem with your argument is that your $x$ in (1) is not an integer (unless you prove it). So even though you have $3cb=(c−b)^{2}(x^3−1)$, it does not imply that $c-b$ divides $3cb$.

Note also that $c-b$ can divide $c$ and $b$ even for coprime $c$ and $b$, contrary to what you say. For example, whenever $c-b=1$, it divides $c$ and $b$, while $c$ and $b$ are necessarily coprime.

Answer 3

Suppose a^3 + b^3 = c^3 with (a,b,c) = 1. Let (1) a + b = c + x. Cubing, (2) a^3 + 3a^2(b) + 3(a)b^2 + c^3 = c^3 + 3c^2(x) + 3(c)x^2 + x^3.
Subtracting a^3 + b^3 = c^3, (3) 3ab(a + b) = 3cx(c + x) + x^3. Transposing and substituting (1), (4) 3(a + b)(ab - cx) = x^3. Therefore, x = 3r, r>0. Substituting x+3r into (4), (5) 3(a + b)(ab - 3cr) = 27r^3, and dividing by 3, (6) (a + b)(ab - 3cr) = 9r^3. The left side is divisible by r^3, but r cannot divide both factors. If so, we would have: (7) a + b = re and ab - 3cr = rf, (e,f) = 1. now if p is a prime factor of r, the second of these equations shows that p would divide ab, and therefore, p|a or p|b. Then referring to the first of these two equations, we have that p divides a and b, contrary to (a,b)= 1. Therefore,r can have no prime factors, so r = 1. Equation (6) becomes: (8) (a + b)(ab - 3c) = 9. As before, the left

Answer 4

In May 1995, in de Annals of Mathematics, a proof of Andrew Wiles is published that there is no solution with integers for A^n + B^n = C^n for n>2. See https://nl.wikipedia.org/wiki/Andrew_Wiles.