Simple method to prove $a^3+b^3=1$ has no integer solutions if $ab\neq 0$ Any simple method to prove $a^3+b^3=1$ has no integer solutions if $ab\neq 0$ that does not involve Fermat's last theorem?
 A: $$1 = a^3+b^3 = (a+b)(a^2-ab+b^2) = (a+b)((a+b)^2-3ab)$$
implies $a+b=\pm 1$ and so $(a+b)^2 - 3ab = 1 - 3ab \not\in \{-1,1\}$ if $ab \ne 0$
A: Well $a^3 + b^3 = (a+b)(a^2 -ab + b^2)=1$
The only factors of $1$ are either $1\cdot 1$ or $-1 \cdot (-1)$.
So we must have $a+b = a^2 - ab + b^2 =\pm 1$.
And that just won't work unless $ab=0$.
It would follow that $b= \pm 1 -a$
And therefore  $a^2 - a(\pm 1 - a) + (\pm 1 - a) = \pm 1$ so
$a^2+a^2 \mp a -a \pm 1 = \pm 1$
SO either $2a^2 -2a+1=1$ or $2a^2 -1 = -1$ so
$2a(a-1) = 0;b= 1-a$ or $2a^2 = 0;b = -1-a$.
So we have either $a=0$ and $b=1$.  Of $a =1$ and $b =0$.  Or $a=0$ and $b =  -1$.
In all three cases we must have $ab = 0$.
A: Let $ab\neq 0$
$$a^3+b^3=(a+b)(a^2-ab+b^2)=1$$ Therefore $a,b$ cannot have the same signs, then $$ab < 0$$ Hence $$a^2-ab+b^2>1$$ which is a contradiction.
A: Tedious (and therefore somewhat inferior) alternative approach that focuses on the minimum difference between distinct cubes.
Prove that the following is impossible: 

*

*$a,b \in \Bbb{Z}$ 

*$a \neq 0 \neq b.$ 

*$a^3 + b^3 = 1.$
For $c,k \in \Bbb{Z},~$ let $f(c,k) = (3c^2k + 3ck^2 + k^3).$
$\underline{\text{Lemma 1}}$ 
$(c,k \in \Bbb{Z^+}) ~\implies ~f(c,k) \neq 1.$ 
Proof : 
$f(1,1) = 7.$ 
$(c \geq 1) ~\implies~ (c^2 \geq 1).$ 
$(k \geq 1) ~\implies~ (k^2 \geq 1 ~~~\text{and}~~~ k^3 \geq 1).$ 
Therefore,
$~~(c \geq 1 ~~~\text{and}~~~ k \geq 1) ~\implies~ f(c,k) \geq f(1,1).$
$\underline{\text{Lemma 2}}$ 
$(c,k \in \Bbb{Z^+}) ~\implies ~ \{[(c+k)^3 - c^3] \neq 1\}.$ 
Proof : 
$[(c+k)^3 - c^3] = f(c,k)$.
Invoke Lemma 1.

If $a,b$ both positive, then the minimum value of $a^3 + b^3$ is $2$. 
If $a,b$ both negative, then the maximum value of $a^3 + b^3$ is $-2$. 
Therefore, $~~~~a \in \Bbb{Z^+},~ b \in \Bbb{Z^-},~~~~$ or vice versa.
Since the equation in dispute, $1 = a^3 + b^3$, is symmetric with respect to $a$ and $b$, 
without loss of generality,
$~~a \in \Bbb{Z^+},~ b \in \Bbb{Z^-}$
Let $c = -b \implies$

*

*$c \in \Bbb{Z^+}~~$ and

*$1 = a^3 + b^3 = a^3 - c^3$.

This implies that $a > c.$ 
Therefore, $~~~\exists k \in \Bbb{Z^+}~~~$ such that $~~~a = (c + k)$. 
Invoke Lemma 2.
