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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.


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


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 $~c,~b$)

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I don't see why $c-b$ divides $3cb$... – Nishant Jul 30 '14 at 2:05
That this has no non-zero solutions was proved by Euler, and perhaps earlier by Fermat. The proof is more complicated than the proof for $x^4+y^4=z^4$. – André Nicolas Jul 30 '14 at 2:17
@frogeyedpeas $c-b$ is factor of $a$. – miket Jul 30 '14 at 5:59
I don't see the point your trying to make, your single case merely reveals "Sometimes the fraction is indeed an integer" whereas you are right now attempting to prove to everyone that at all times, $\frac{a^3}{(c-b)^3}$ is an integer divisible by $c-b$ whenever $a^3 + b^3 = c^3$, re-read this comment and see if you understand what I mean – frogeyedpeas Jul 30 '14 at 6:53
To hit your edit: you haven't shown that of necessity there's no integer solution. You're right that the algebraic expression $\frac16(\sqrt{3}\sqrt{4a^3-1}+3)$ isn't a polynomial in $a$, but you still haven't shown any reason why it can't be an integer. (For a simpler example, $\sqrt{2x^2-1}$ isn't a polynomial expression, but that doesn't mean that it can't be an integer - take $x=12$, for instance.) – Steven Stadnicki Aug 1 '14 at 3:18

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.

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$c−b\ $ divides $\ 3cb\ $ related problem can be solved.$\ c-b=1\ $ this is a problem. – miket Jul 31 '14 at 8:59
@miket: $c-b=1$ was only mentioned as a quick example, there are of course other examples where $c$ and $b$ are coprime, yet $c-b$ divides $cb$. Even if $c-b$ does not divide $cb$, it is not true that $c-b\mid 3cb$ implies $c-b\mid 3$. This was only true if $c-b$ were relatively prime to $cb$, which is a much stronger relation than not dividing $cb$. – anonymous Jul 31 '14 at 23:05
And, as I mentioned, the first serious problem with your reasoning is that you rely on an unproven statement, namely that $x$ is an integer. – anonymous Jul 31 '14 at 23:08
There's two problmes left.See update. – miket Aug 1 '14 at 1:58
@anonymous To be fair, the only case in which $c-b$ can divide coprime $b$ and $c$ is when $c-b=(\pm)1$. – Steven Stadnicki Aug 1 '14 at 3:58

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.

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Yes. $\ c-b\ |\ a^3\ $there's three case: $\ 1.\quad c-b \nmid a,\quad\ 2. \quad\ c-b \mid a,\ \quad3. \quad\ c-b \nmid a,\ \ c-b \ $ contains other factors of $a$ – miket Jul 31 '14 at 8:56
$c=9,b=1,a=2$ is simpler. Or $c=11,b=3,c=2$. – user236182 Dec 11 '15 at 13:45

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