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.
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.