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Legendre conjectured that for $p, q, r, s \in \mathbb{N}$ it is impossible that $$\biggl(\frac{p}{q}\biggl)^3+\biggl(\frac{r}{s}\biggl)^3=6$$ for any $p, q, r, s$.

Eventually Henry Ernest Dudeny found a counterexample, where $p=17, q=21, r=37, s=21$: $$\biggl(\frac{17}{21}\biggl)^3+\biggl(\frac{37}{21}\biggl)^3=\frac{4913}{9261}+\frac{50653}{9261}=6.$$

Does the original conjecture hold up if we assume that none of $p, q, r, s$ repeat? That is, none of the four numbers can equal each other?

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    $\begingroup$ So ... if there is a prime that divides one of the denominators more times than the other, then your sum cannot be an integer... Try that. $\endgroup$ – GEdgar Jul 8 '18 at 18:48
  • $\begingroup$ @GEdgar I assume you mean a case where the fractions are reduced as far as they can be? So doesn’t that mean it’s impossible? $\endgroup$ – DonielF Jul 8 '18 at 23:31
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The given equation is equivalent to $$p^3s^3+q^3r^3=6q^3s^3$$ If we assume that $p$ and $q$ are coprime as well as $r$ and $s$ (otherwise the counterexample can easily be modified such that all numbers are distinct) , we get $q^3|s^3$ because of $q^3|p^3s^3$ and $s^3|q^3$ because of $s^3|q^3r^3$. This implies $s^3=q^3$ and therefore $s=q$

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    $\begingroup$ +1... It is easily shown that $p/q\ne 1\ne r/s,$ so if we drop the condition that $p/q $ is in lowest terms we can replace $(p,q)$ by $(pk,qk)$ where $k\in \Bbb N$ with $\min (pk,qk)>\max (r,s)$ and get $p,q,r,s$ to be $4$ distinct numbers. $\endgroup$ – DanielWainfleet Jul 9 '18 at 13:26

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