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Prove that if $m, n, d\in\mathbb{Z^+}$ and $d \mid m^2n+1, d \mid mn^2+1$ then $d \mid m^3+1, d \mid n^3+1$

My attempt:

So I said that from this we can safely say that $$m^2n\equiv-1\pmod{d}, mn^2\equiv-1\pmod{d}$$ This means that: $$mn(m)\equiv mn(n)\pmod{d}$$ Now if $\gcd(mn, d)=1$ then we will get that $$m\equiv{n}\pmod{d}$$ Now in our original equation we will get $$m^2n+1\equiv0\pmod{d}, mn^2+1\equiv0\pmod{d}$$ $$\therefore m^3+1\equiv0\pmod{d}, n^3+1\equiv0\pmod{d}$$ But what if $\gcd(mn,d)\gt{1}$, what will I have to do? And is that even possible in the first place? Thank you in advance.

Martin Sleziak
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user587054
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    $\gcd(mn,d)$ has to be $1$, because if they had a common factor, that factor would divide both $mn^2+1$ and $mn^2$. – Wojowu Jun 17 '19 at 10:16
  • $\bbox[4px,border:1px solid #c00]{ a,b\equiv 1, \Rightarrow, a^2/b\equiv 1}\ \ $ Put $\ a = -m^2n,, b = -mn^2,\ $ so $\ {-}m^3\equiv a^2/b \equiv 1\ \ \ \small\bf QED\ \ $ – Bill Dubuque Jan 26 '23 at 18:53
  • Or we can eliminate $,n,$ via $,\color{#c00}{n\equiv -m^{-2}},$ so $,-1\equiv m\color{#c00}{n^2}\equiv m^{-3},$ so $,m^3\equiv -1$. This is a special case of using nonmonic ("fraction free") division in the Euclidean algorithm as explained here, i.e. $,(m^2n+1,mn^2+1) = (m^2n+1,m^3(mn^2+1)) = (m^2n+1,(-1)^2+m^3),,$ i.e. before eval'ing $f(n) = mn^2+1$ at $,n\equiv -m^{-2}$ we first scale it by the unit $,m^3,$ to elminate all "fractions". – Bill Dubuque Jan 26 '23 at 20:38

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Let $p$ be any prime divisor of $d$, and let $k$ be the maximum natural number such that $p^k \mid d$. Observe that: $p^k \mid d \mid mn(m-n)$, and $m^2n+1 = m^2(n-m+m)+1 = m^3+1 + (n-m)m^2$. This means that if $p^k \nmid (n-m)\implies p^k \mid mn\implies p \mid 1$, contradiction $p$ being a prime. Thus $p^k \mid (n-m)$,and this means $p^k \mid m^3+1$. Thus $d \mid (m^3+1)$, and similarly $d \mid (n^3+1)$. Note that if $p^r \mid (n-m)$, and $p^s \mid mn$ with $r+s = k$, then $p \mid m^2n \implies p \mid 1$, contradiction $p$ being a prime.

DeepSea
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