<s>[INST] If $(a-1),a,(a+1)$ are consecutive positive integers, $ (a+1)^3 \neq a^3 + (a-1)^3$ I had to prove the following statement: 

If $(a-1),a,(a+1)$ are consecutive positive integers, $(a+1)^3 \neq a^3 + (a-1)^3$

My attempt at the solution was to first expand each side to get 
$$a^3 + 3a^2 + 3a + 1 \neq 2a^3 - 3a^2 +3a - 1\\
      0 \neq a^3 - 6a^2 - 2$$
However, $a^3 - 6a^2 - 2$ does hit the $x$-axis at $a = 6.0546$.

Does that mean that the statement is incorrect?

 [/INST] Suppose $n^3-6n^2 = 2$ for some integer $n$. It is easy to check that $n$ cannot be odd. Hence $n = 2k$ for some integer $k$. We have $8k^3-24k^2 = 2$, or equivalently, $k^3-3k^2 = \frac{1}{4}$, which is a contradiction since $k$ was supposed to be an integer.
</s>