<s>[INST] Prove: If $a$, $b$, and $c$ are consecutive integers such that $a< b < c $ then $a^3 + b^3 \neq c^3$. 
Prove: If $a$, $b$, and $c$ are consecutive integers such that $a<  b <  c $ then $a^3 + b^3  \neq  c^3$.

My Attempt: I start with direct proof.
Let $a,b,c$ be consecutive integers and  $a<  b <  c $, there exists a integer $k$ such that $a=k, b=k+1, c=k+2$. Then $a^3+b^3=k^3+(k+1)^3=k^3+(k^3+3k^2+3k+1)$ and $c^3=(k+2)^3=k^3+6k^2+12k+8=(k^3+3k^2+3k+1)+(3k^2+9k+7)$. Hence, we have $k^3+(k^3+3k^2+3k+1)\neq (k^3+3k^2+3k+1)+(3k^2+9k+7)$ which implies $a^3+b^3\neq c^3$.
Does my proof valid ? And, is use Contradiction easier? If not, can anyone give me a hit or suggestion ? 
Thanks
 [/INST] $(k-1)^3+k^3=(k+1)^3\\\implies k^3=(k+1)^3-(k-1)^3\\\implies k^3=(k+1-k+1)(k^2+2k+1+k^2-1+k^2-2k+1)\\\implies k^3=2(3k^2+2)$
Let $p$ be a prime that divides $k$.
If $p$ be odd then $k^3=2(3k^2+2)\implies p\mid3k^2+2\implies p\mid 2$
Hence $k$ can't have an odd prime divisor. Therefore $k=2^t$. $(?)$
Now assume that $t>0\implies t\ge 1$
Then we have,
$2^{3t}=2(3\cdot2^{2t}+2)\implies 2^{3t-1}=3\cdot 2^{2t}+2\implies2^{3t-2}=3\cdot 2^{2t-1}+1\implies t=\dfrac{1}{2}\ (?)$
Which implies $t=0$. But putting $t=0$ the equation is not satisfied. Hence the conclusion follows.
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