# Show that the quantities are natural numbers

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

• 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 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]$$

• If the hint is too terse I can elaborate. Let me know. – Bill Dubuque Oct 8 '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 9 '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 9 '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 9 '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 9 '18 at 13:48

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?

• Why does $d=gcd(x,z)$ divide $w$ and $y$ ? @CarlSchildkraut – Evinda Oct 8 '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 8 '18 at 18:07
• Ok, I see.. But how do we proceed after considering that gcd(x,z)=1 ? – Evinda Oct 8 '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 8 '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 8 '18 at 18:55