$\frac{a^{3}+1}{b+1}+\frac{b^{3}+1}{a+1}$ an integer $\Rightarrow \frac{a^{3}+1}{b+1}$ and $\frac{b^{3}+1}{a+1}$ are integers.

Question

I want to show that if the natural numbers $a,b \in \mathbb{N}$ are such that $\frac{a^3+1}{b+1}+\frac{b^3+1}{a+1} \in \mathbb{N}$, then, necessarily, $\frac{a^3+1}{b+1} \in \mathbb{N}$ and $\frac{b^3+1}{b+1} \in \mathbb{N}$.

I have thought the following.

We are given that $\frac{a^3+1}{b+1}+\frac{b^3+1}{a+1} \in \mathbb{N}$.

This means that $\frac{a^3+1}{b+1}+\frac{b^3+1}{a+1}=k, \text{ for some } k\in \mathbb{N}$.

We also have that $b+1 \mid a^3+1$ and $a+1 \mid b^3+1$, right?

So does it suffice to reject the cases that $a^3+1=k_1(b+1)$ and $b^3+1=k_2 (a+1)$ for negative $k_1, k_2$ ? If so, then we pick all the possible combinations and want to get a contradiction from the fact that $\frac{a^3+1}{b+1}+\frac{b^3+1}{a+1} \in \mathbb{N}$, or not?

Or do we show somehow else that $\frac{a^3+1}{b+1} \in \mathbb{N}$ and $\frac{b^3+1}{b+1} \in \mathbb{N}$ ?

Answer 1

Hint $\ r\! +\! s,\, rs \in \Bbb Z\,\Rightarrow\, r,s \in \Bbb Z\ $ by applying the Rational Root Test to $\,(x\!-\!r)(x\!-\!s)\in\Bbb Z[x]$

Answer 2

Hint: You can prove this more general statement:

Let $w,x,y,z$ be positive integers so that

$$\frac{w}{x}+\frac{y}{z}\in\mathbb{N}$$

and $z|w,x|y$. Then

$$\frac{w}{x},\frac{y}{z}\in\mathbb{N}.$$

To do so, consider $d=\gcd(x,z)$, which must divide each of $w$ and $y$, so by dividing each variable by $d$ it suffices to consider the case where $\gcd(x,z)=1$. Can you finish from here?