Proving there is only one set of positive integer solutions I am looking to prove there is no perfect cube root to any thing of the form $3k^2 + 3k + 1$. As has been discussed previously, we know this must be true by invoking Fermat's last theorem, because $3k^2 + 3k + 1 = (k+1)^3 - k^3$, and after re-arrangement we can simply state there are no integer solutions to $(k+1)^3 = w^3 + k^3$. But this might be a little over the top, and I am convinced there must should be much more trivial ways to show this. One approach I took was this to let $w = k-d$ where $d$ is an integer. Then, $3k^2 + 3k + 1 = k^3 -3dk^2 + 3kd^2 -d^3 $. After re-arrangement, this gives 
$k^3 -3k^2(d+1) + 3k(d^2 -1) - (d^3 + 1) = 0$
The trivial solution is to note that $k = 0$ eliminates all $k$ terms, and then one is left with $d = -1$. This makes sense, as it forces one of the cube terms to zero and doesn't violate Fermat's theorem in anyway. But the question is, can I tell by inspection alone whether this is the only non-negative integer value of $k$ that solves the equation? 
 A: I am not entirely sure if this answer is valid, but let me try explain my reasoning. We wish to show there's no rational form for the cube root of $3k^2 + 3k + 1$ when $k > 0$. We know this is true (it's the same as writing $(k-1)^3 - k^3$ which by Fermat's last theorem has no integer solutions when $k>0$, but we want to prove it a more straight-forward way. Let's assume there is a perfect cube root $w$, in which case 
$3k^2 + 3k + 1 = w^3$
We can also write $w = k +d$ where d is a negative or positive integer, and thus we have  
$3k^2 + 3k + 1 = (k+d)^3$.
Noting that $(k + d)^3 = k^3 + 3k^2d + 3kd^3 + d^3$, we note that the only way the $k$ and $k^2$ terms could match the original equation is if $d = 1$. And it follows from this that for the polynomials to match, $k^3 = 0$ and so $k =0$, and there are no other integer solutions possible. 
As $k \geq 0$, essentially, there's no clever integer combination that eliminates precisely $k^3$ from the binomial theorem, and thus only $k=0$ is an integer solution.
Note:I'm not sure this is perfect, and there must also be more elegant ways to solve it. If the $k \geq 0$ condition was relaxed, for example, Mathematica also finds $(k,d)= (-1,2)$ - if anyone has a more sophisticated answer, I'm happy to hear it!   
