Find all integers $a$ and $b$, so that $ab$ is a factor in $a^5+b$

Question

Find all integers $a$ and $b$, so that $ab$ is a factor in $a^5+b$

I know that $b=-a^5$ is a solution, but are there any more solutions?

Answer 1

Hint: Let's say we have $ab|a^n+b$ for some fixed $n$. Then $a|ab|a^n+b$, so $a|b$. Thus, let $b=ka$. Then, we need

$$ka^2\big|a^n+ka$$

$$ka\big|a^{n-1}+k$$

We can reduce this down to $ka|1+k$. Can you solve it from here, and use this to find all solutions to $ab|a^5+b$.

Answer 2

Hint $\ abc = a^5\!+\!b \!\iff\! b(ac\!-\!1) = a^5\!\iff\! b = \dfrac{a^5}{ac\!-\!1}.\,$ It's reduced by $\,\gcd(a,ac\!-\!1)=1$ hence $\,b\in\Bbb Z\iff\, ac\!-\!1 = \pm1\iff \ldots$

Answer 3

So $$ab\mid a^5+b \implies a\mid a^5+b \implies a\mid b$$

so $b=ca$. Now we have $$aca\mid a^5+ca \implies ac\mid a^4+c \implies a\mid c$$

so $c=da$. Now we have $$ada\mid a^4+da \implies ad\mid a^3+d \implies a\mid d$$

so $d=ea$. Now we have $$aea\mid a^3+ea \implies ae\mid a^2+e \implies a\mid e$$

so $e=fa$. Now we have $$afa\mid a^2+fa \implies af\mid a+f \implies a\mid f$$

so $f=ga$. Now we have $$aga\mid a+ga \implies ag\mid 1+g \implies g\mid 1$$

so $g=1$ and $a\mid 2$ so $a\in \{1,-1,2,-2\}$ and $b...$