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?
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asked Jan 26, 2020 at 17:33
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3$\begingroup$ Note that $\tfrac{a^2+ab+b^2}{a+b}=a+\tfrac{b^2}{a+b}$, so $(a,b)=(d-b,b)$ where $d$ is any divisor of $b^2$. $\endgroup$– ServaesJan 26, 2020 at 17:37
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1$\begingroup$ $\frac {a^2 + ab + b^2}{a+b} = \frac {(a + b)^2 - ab}{a+b}$ We get an integer when $(a+b)|ab$ $\endgroup$– Doug MJan 26, 2020 at 18:20
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$\begingroup$ It seems that you ask about $a,b\in\mathbb Z$ or $a,b\in\mathbb N$, but it might be good to say so explicitly also in the question. $\endgroup$– Martin SleziakJan 26, 2020 at 18:42
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$\begingroup$ @MartinSleziak I did but someone edited that $\endgroup$– Consider Non-Trivial CasesJan 26, 2020 at 18:43
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3 Answers
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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 $$a=d+k,\qquad b=\frac{k^2}{d}+k.$$
answered Jan 26, 2020 at 18:36
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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:
$(a,b,p)=(15,30,35)$
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A trivial example is for $a,b \in \mathbb{Z}$ such that $a+b=1$
