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 $\frac{b^3+1}{b+1} \in \mathbb{N}$ ?

  • $\begingroup$ You can't go from $\frac{w}{x}+\frac{y}{z}\in\mathbb{N}$ to $x|w$ and $z|y$. You're implicitly assuming that $\frac{w}{x}$ and $\frac{y}{z}$ are integers when you go through that step. $\endgroup$ – Carl Schildkraut Oct 8 '18 at 17:37

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

  • 1
    $\begingroup$ If the hint is too terse I can elaborate. Let me know. $\endgroup$ – Bill Dubuque Oct 8 '18 at 19:43
  • $\begingroup$ Yes, could you please explain it further to me? I haven't understood how we apply the Rational Root Test? @BillDubuque $\endgroup$ – Evinda Oct 9 '18 at 11:36
  • $\begingroup$ Applying the Rational Root test to $(x-r)(x-s)$, we deduce that the only roots are the divisors of $rs$. Right? How does this help? @BillDubuque $\endgroup$ – Evinda Oct 9 '18 at 12:05
  • $\begingroup$ @Evinda RRT implies that if a polynomial with integer coef's is monic (lead coef $= 1$) then every rational root is an integer, because the denominator of any reduced rational root must divide the lead coef. This is a fundamental fact used widely in number theory and algebra. $\endgroup$ – Bill Dubuque Oct 9 '18 at 13:38
  • $\begingroup$ From your last mentioned statement, we have that all the roots of the polynomial $x^2-(r+s)x+rs$ are integers. And we also have that the roots are divisors of $rs$, right? Two possible divisors of $rs$ are $r$ and $s$, and so we have that $r$ and $s$ are integers, right? @BillDubuque $\endgroup$ – Evinda Oct 9 '18 at 13:48

Hint: You can prove this more general statement:

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


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


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?

  • $\begingroup$ Why does $d=gcd(x,z)$ divide $w$ and $y$ ? @CarlSchildkraut $\endgroup$ – Evinda Oct 8 '18 at 18:03
  • $\begingroup$ If we have that $z|w$ and $x|y$, then $d|z|w$ and $d|x|y$ for $d=\gcd(x,z)$. $\endgroup$ – Carl Schildkraut Oct 8 '18 at 18:07
  • $\begingroup$ Ok, I see.. But how do we proceed after considering that gcd(x,z)=1 ? $\endgroup$ – Evinda Oct 8 '18 at 18:36
  • $\begingroup$ Consider what happens if $x\nmid w$ or $z\nmid y$. What would have to be in the denominator of the other term? $\endgroup$ – Carl Schildkraut Oct 8 '18 at 18:47
  • $\begingroup$ Do we then set $w=\lambda x+k$ for $k \neq 0$ ? Then $\frac{w}{x}+\frac{y}{z}=\lambda+\frac{k}{x}+\frac{y}{z}$. Can we get somehow that the latter is not a natural number? Or do we have to do something else? $\endgroup$ – Evinda Oct 8 '18 at 18:55

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