When $\frac {a^3-b^3}{a^2-b^2}$ is an integer?

Question

If $\frac {a^3-b^3}{a^2-b^2}$ is an integer, then supposing $a-b \ne 0$ we have that also$\frac {a^2+ab+b^2}{a+b}$ is an integer.

For which $a, b\in\mathbb Z$, the fraction $\frac {a^2+ab+b^2}{a+b}$ is an integer?

Answer 1

Assuming $a\neq b$, we want that $\frac{a^2+ab+b^2}{a+b}=(a+b)-\frac{ab}{a+b}$ is an integer. The solutions of $$ \frac{ab}{a+b} = k, $$ assuming $a+b\neq 0$, are the solutions of $$ ab-ka-kb = 0, $$ i.e. the solutions of $$ (a-k)(b-k) = k^2, $$ which depend on the couples of divisor/complementary divisor of $k^2$.
In general, for any $d\mid k^2$ we have the solution $$a=d+k,\qquad b=\frac{k^2}{d}+k.$$

Answer 2

A trivial example is for $a,b \in \mathbb{Z}$ such that $a+b=1$

Answer 3

Equation, $(a^2+ab+b^2)=p(a+b)$

where "p' is integral.

while solution given by "Jack D'Aurizio" is nice & since

'OP' needs $(a,b)$ to be integer's there is a

fraction $(k^2/d)$ to be taken care of in his solution.

If instead we take, $(a,b)=[d(k+1),dk(k+1)]$ then we get:

$p=d(k^2+k+1)$

For, $(d,k)=(5,2)$ we get:

$(a,b,p)=(15,30,35)$