1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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

$\endgroup$
  • 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
2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.