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Prove for all $a, b, d$ integers greater than zero, that when $d$ is a divisor of both $a^2b+1$ and $b^2a+1$ then $d$ is also a divisor of both $a^3+1$ and $b^3+1$

There is a trivial case when $a = b$, but I found that there are more solutions, for example: $a = 1, b = 3, d = 2$. I tried to use equation $a^3+1=(a+1)(a^2-a+1)$ but I am not sure how to progress.

Eric Snyder
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Krzysztof
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  • $\bbox[4px,border:1px solid #c00]{ c,e\equiv 1, \Rightarrow, c^2/e\equiv 1}\ \ $ Put $\ c = -a^2b,, e = -ab^2,\ $ so $\ {-}a^3\equiv c^2/e \equiv 1\ \ \ \small\bf QED\ \ $ – Bill Dubuque Jan 26 '23 at 18:51
  • Or we can eliminate $,b,$ via $,\color{#c00}{b\equiv -a^2},$ so $,−1\equiv a\color{#c00}{b^2}\equiv a^{-3},$ so $,a^3\equiv -1.,$ This is a special case of using nonmonic ("fraction free") division in the Euclidean algorithm as explained here, i.e. $,(a^2b+1,ab^2+1)=(a^2b+1,a^3(ab^2+1))=(a^2b+1,(−1)^2+a^3),,$ i.e. before eval'ing $,f(b)=ab^2+1,$ at $,b\equiv −a^{−2},$ we first scale it by the unit $,a^3,$ to elminate all "fractions". – Bill Dubuque Jan 26 '23 at 20:43

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Since $d \mid a^2\ b+1$ and $d \mid a\ b^2+1$ then $d \mid (1-b\ a^2)(a^2\ b+1)+a^3(a\ b^2+1) = a^3+1$

jjagmath
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  • Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Jan 26 '23 at 17:52
  • Said equationally it's that below. But it's easier to manipulate units (see my comment on the question) $$\bmod d!:,\ \overbrace{1\equiv \color{#c00}{a^4b^2}}^{!!!!\textstyle (-1\equiv \color{#c00}{a^2b})^{\color{#c00}2}}!!!\equiv \underbrace{a^3(\color{#0a0}{ab^2})\equiv -a^3}_{\textstyle \color{#0a0}{ab^2}\equiv -1!}: \Rightarrow\ d\mid a^3+1\qquad$$ – Bill Dubuque Jan 26 '23 at 18:44