Suppose $(a^2+3ab+3b^2-1)|(a+b^3)$. Find all integer solutions for $a$ and $b$ Let $a$ and $b$ are positive integers. Suppose $(a^2+3ab+3b^2-1)|(a+b^3)$.
Show $k^3 | (a^2+3ab+3b^2-1)$, where $k>1$ and $k$ is some positive integer. Further, find all possible integer solutions for $a$ and $b$.
My attempt is to let $(a^2+3ab+3b^2-1)t=(a+b^3)$, so we got a quadratic equation with respect to $a$, use Vieta's formula we got the sum for two roots $a_1+a_2=-3b+\frac{1}{t}$. If we assume both roots are integers, then $t=1$. Further discussion gives: $a=n^3-3n+1, b=n^2+n-1$ where $n\ge2$
Question is: $a_1+a_2=-3b+\frac{1}{t}$, maybe only one root is integer and another root is rational, which means $t$ could be $2, 3, 4,...$, so how to find integer solutions for other cases of $t$, or how to exclude other cases?
 A: This is not an answer, but it may shed some light on the problem, and it's too long for a comment.
Let's look at $$(a^2+3ab+3b^2-1)t=a+b^3$$ For any given nonzero integer $t$, we can make some substitutions to get this into the form $y^2={\rm a\ cubic\ in\ }x$. Such an equation is, in general, an elliptic curve, also known as a curve of genus one, and a theorem of Siegel states that it has only finitely many solutions in integers $x,y$ (where "finitely many" could be none at all). Given an elliptic curve, it's not all that easy to find all the integer solutions, or even to determine whether it has any. Usually, some high-powered math is involved. As $t$ varies, we have a family of elliptic curves, and it's generally even harder to find all the solutions, or to show there are none.
Now, $t=1$ is an exception. Let $f(a,b)=a^2+3ab+3b^2-1-a-b^3$, and let $g(a,b)=f(a-1,b+1)=a^2+3ab-b^3$. Then $g(0,0)=0$, and both the partial derivative of $g$ with respect to $a$ and the partial derivative of $g$ with respect to $b$ vanish at $(0,0)$, so there is a singular point at the origin, and $g(a,b)=0$ is of genus zero, and not an elliptic curve. A line through the origin with slope $r$, $a=br$, intersects the graph of $g(a,b)=0$ at $(0,0)$ and at $(a,b)=(t^3+3t^2,t^2+3t)$, giving rise to an infinity of integer solutions – translating back to $f(a,b)$, this gives the infinite family of solutions OP found.
Whether there are solutions for other values of $t$ is, as already noted, probably a harder problem because they will probably be sporadic and not come in nice infinite families. In addition to running a computer on it to see whether there are any smallish solutions, it may be a good idea to post to MathOverflow, where users should have more to say about the mathematics of families of elliptic curves.
