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

Bill Dubuque
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Evinda
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  • 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. – Carl Schildkraut Oct 08 '18 at 17:37

2 Answers2

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

Bill Dubuque
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    If the hint is too terse I can elaborate. Let me know. – Bill Dubuque Oct 08 '18 at 19:43
  • Yes, could you please explain it further to me? I haven't understood how we apply the Rational Root Test? @BillDubuque – Evinda Oct 09 '18 at 11:36
  • 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 – Evinda Oct 09 '18 at 12:05
  • @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. – Bill Dubuque Oct 09 '18 at 13:38
  • 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 – Evinda Oct 09 '18 at 13:48
  • Or is it possible that we have a solution other than $r$ or $s$ and so my above statemnt is wrong? – Evinda Oct 09 '18 at 14:03
  • @Evinda $r,s$ are roots of $(x-r)(x-s)= x^2-(r+s)x + rs$. Choosing $r,s$ as your summands completes the proof since they have integer sum and product. – Bill Dubuque Oct 09 '18 at 14:04
  • I see... Thanks a lot :) @BillDubuque – Evinda Oct 09 '18 at 15:55
  • @Evinda Great! Please feel welcome to ask further questions if need be. It is not unusual to be puzzled when one first encounters these pretty applications of RRT. – Bill Dubuque Oct 09 '18 at 16:04
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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?

Community
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Carl Schildkraut
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  • Why does $d=gcd(x,z)$ divide $w$ and $y$ ? @CarlSchildkraut – Evinda Oct 08 '18 at 18:03
  • If we have that $z|w$ and $x|y$, then $d|z|w$ and $d|x|y$ for $d=\gcd(x,z)$. – Carl Schildkraut Oct 08 '18 at 18:07
  • Ok, I see.. But how do we proceed after considering that gcd(x,z)=1 ? – Evinda Oct 08 '18 at 18:36
  • Consider what happens if $x\nmid w$ or $z\nmid y$. What would have to be in the denominator of the other term? – Carl Schildkraut Oct 08 '18 at 18:47
  • 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? – Evinda Oct 08 '18 at 18:55
  • @Evinda Above is a special case of the fact that rationals who sum and product are integers must themselves be integers, by RRT = Rational Root Test, see my answer. RRT is the master theorem for problems like this. Learn it well. – Bill Dubuque Oct 08 '18 at 19:27
  • @BillDubuque I haven't really understood how we can deduce from the rational root test that $r,s \in \mathbb{Z}$. – Evinda Oct 09 '18 at 12:08
  • @CarlSchildkraut I am also wondering how can we finish the proof with the way you used. Can you give me a hint how we continue? – Evinda Oct 09 '18 at 12:08
  • @Evinda Assume $\frac{w}{x}$ reduces to $\frac{m}{n}$, where $n>1$. Then $\frac{y}{z}$ must reduce to $\frac{m'}{n}$, where $n|m+m'$. However, we have that $n|x$ and $n|z$, so... – Carl Schildkraut Oct 09 '18 at 13:50
  • Why does then $\frac{y}{z}$ have to be reduced to $\frac{m'}{n}$ where $n \mid m+m'$ ? – Evinda Oct 09 '18 at 15:59
  • @Evinda because it is an integer minus $m/n$. – Carl Schildkraut Oct 09 '18 at 16:10
  • Ah, I see. So since $n \mid x$ and $n \mid z$, we also have that $n \mid gcd(x,z)=1 \Rightarrow n=1$, which is a contradiction. Right? – Evinda Oct 09 '18 at 17:58
  • @Evinda Yes, that's correct! – Carl Schildkraut Oct 09 '18 at 20:45