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Let $k$ and $m$, where $k>m$, be positive integers such that $$k^3-m^3 \vert km(k^2-m^2).$$ Prove that ($k-m)^3>3km$

I established that $k^3-m^3<km(k^2-m^2)$, which gives $k^2+m^2+km<km(k+m)$. By AMGM $3km<k^2+m^2+km$.

Could someone give me a hint? I'm stuck but I would like to solve it by myself.

Bill Dubuque
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Arthr
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  • Have you tried using the general solution in the answer that I provided to your earlier related question, in particular using inequalities based on the lower limits of certain values? Note I've confirmed that this does work. However, I won't post an answer here since you've stated that you would like to solve it yourself. – John Omielan Jan 06 '24 at 18:06
  • @JohnOmielan I think I've got it. I posted my solution below. Could you please confirm my solution? Thank you for your help – Arthr Jan 06 '24 at 18:42
  • You're welcome for my assistance. It's great that you tried to get a solution using the forms for $k$ and $m$ that solve $k^3 - m^3 \mid km(k^2 - m^2)$. Unfortunately, as I stated in my answer, the expressions for $k$ and $m$ you're using are only a special case. Nonetheless, you've got the right general idea with what you've shown in your answer. I suggest you now try to use the full general form I give in my $(5)$ to reach the same conclusion. – John Omielan Jan 06 '24 at 18:47
  • @JohnOmielan I think now it's right – Arthr Jan 06 '24 at 19:05

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I think I've got now a solution. With the expressions for k,and m from this question

Set $k=dk_1$, $m=dm_1$ and $d=n(k_{1}^2+k_1m_1+m_{1}^2)$.
Then the inequality becomes:
$3d^2k_1m_1<((k_1-m_1)(d))^3$

Which is equivalent to
$3k_1m_1<(k_1-m_1)^3d$

Since by AmGm $3k_1m_1\leq3nk_1m_1<d$ and $(k_1-m_1)>0$ the inequality holds.

Arthr
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  • Good work. This is basically correct, but there are a few minor issues. We have $3k_{1}m_{1} \le 3nk_{1}m_{1}$, since $n\ge 1$, not $3k_{1}m_{1} \lt 3nk_{1}m_{1}$, and also this is not due to the AM-GM inequality. Also, FYI, instead of using the AM-GM inequality to show $3k_{1}m_{1} \lt k_1^2+k_{1}m_{1}+m_1^{2}$, another method is to use $(k_1-m_1)^2\gt 0;\to;k_1^2+2k_{1}m_{1}+m_1^2\gt 0;\to;k_1^2+k_{1}m_{1}+m_1^2\gt 3k_{1}m_{1}$. – John Omielan Jan 06 '24 at 19:18
  • FYI, here's the way I solved it myself. I used $(k-m)^3-3km=d^2(d(k_1-m_1)^3-3k_{1}m_{1})\ge d^2(d-3k_{1}m_{1})\ge d^2(k_1^2+k_{1}m_{1}+m_1^2-3k_{1}m_{1})=d^2(k_1^2-2k_{1}m_{1}+m_1^2)=d^2(k_1-m_1)^2\gt 0$ – John Omielan Jan 06 '24 at 19:22
  • Thank you! Could you maybe recommend some resources for elementary number theory? – Arthr Jan 06 '24 at 19:22
  • You're welcome. First, note I made a typo in my earlier comment in that my $k_1^2\color{red}{+}2k_{1}m_{1}+m_1^2\gt 0$ should be $k_1^2\color{red}{-}2k_{1}m_{1}+m_1^2\gt 0$ instead. Regarding resources for elementary number theory, I don't have any specific recommendations as I've not used any such material myself recently. Nonetheless, a site search gives several potentially useful questions to check, e.g., ... – John Omielan Jan 06 '24 at 19:33
  • (cont.) Very elementary number theory and combinatorics books. and which texts on number theory do you recommend?. They provide specific recommendations, links to other site questions, and external resources, such as the AoPS, i.e., Art of Problem Solving. Note the AoPS is also a quite good resource for high school math olympiad problems and solutions, where some of them involve challenging elementary number related questions. – John Omielan Jan 06 '24 at 19:37