Can both $n+3\; \text{and}\; n^2+3$ both be cubic number at same time? Where $n$ is an integer number. Not necessarily positive.
I tried writing $x^3 = n+3$ and expressing $n^2+3$ in terms of $x$. I found $x^6 -6x^3+12$ but this doesn't help. How do I prove this?
On
Slightly overkill, but if $n + 3$ and $n^2 + 3$ are both cubes, then so is their product, and so
$$ (n + 3)(n^2 + 3) = n^3 + 3n^2 + 3n + 9 = (n + 1)^3 + 2^3 $$
would be a cube. But as is well known, the only possible solutions to this can occur when one of the cubes is $0$, and so we have that either $n = -3$ or $n = -1$, and we can verify that neither of these yield solutions.
Your question is asking if there is $x,a \in \mathbb{Z}$ such that $$x^6-6x^3+12=a^3$$ Note that if we have $x \ge 2, x \le -2$, then we have that $$(x^2-1)^3 = x^6-3x^4+3x^2-1 < x^6-6x^3+12=a^3$$ And also $$(x^2+1)^3 =x^6+3x^4+3x^2+1 > x^6-6x^3+12=a^3$$ This give us $x^2-1<a<x^2+1$, thus forcing $a$ to be $x^2$. However, as $x^3 \neq 2$ for any integer $x$, this is a contradiction.
The only cases left are $x=-1, 0,1 $, which can be manually checked to be never be cubes.