I have come across a particular Diophantine equation that I cannot seem to crack. I think there might be a way to factor the left hand side, but beside that approach I cannot think of any other ideas.

"Find all solutions $(m,n)$ to the Diophantine equation $m^3-5m+10=2^n.$"

Does anyone have any tips or suggestions to generate the desired solutions, and since this is an Olympiad style problem, provide a complete proof?

  • $\begingroup$ Tip: You have to put $ signs around the MathJax for it to get formatted. $\endgroup$ – saulspatz Apr 5 '18 at 1:45
  • $\begingroup$ I'm going to guess that $m=2,n=3$ is the only solution. This is just based on a quick computer experiment. $\endgroup$ – saulspatz Apr 5 '18 at 1:52
  • $\begingroup$ I thought it might have to do with the lifting theorem but the best I've been able to do is prove that $m$ must be even, which isn't a great deal of help. $\endgroup$ – saulspatz Apr 5 '18 at 4:33
  • $\begingroup$ Actually, that isn't really so. What I showed is that for each power $2^k$ there is exactly one solution to $m^3-5m+10\equiv 0\pmod{2^k}$ and when $k\ge 2, m$ must be even. Still, I don't see how to make any further progress. $\endgroup$ – saulspatz Apr 5 '18 at 6:06

This was a nice diophantine equation that easily breaks via mod bashing and bounding.

Consider the LHS mod 7. We have $LHS \in \{0,1,3,5,6\}$. However, $2^n \in \{1,2,4\}$, so $n$ must be divisible by $3$. Let $p=2^{\frac{n}{3}}$.

We now have $m^3-5m+10=p^3 \iff 5m-10 = m^3-p^3$. This is a great candidate for bounding since the difference between consecutive cubes is a quadratic, and here we have only a linear difference.

For $m \geq 3$, we have $3m^2-3m+1 > 5m-10 > 0$, so $m^3 - 3m^2 + 3m -1 = (m-1)^3 < m^3 - 5m + 10 = p^3 < m^3$, which is impossible. Therefore, $m \leq 2$. If $m \leq -3$, then $m^3-5m+10 <0$, so we don't have solutions here.

Thus, it suffices to check $m=-2,-1,0,1,2$, which yields the only solution as $m=2$, $n=3$.

Thus, the only solution in $(m,n)$ is $(2,3)$.

  • $\begingroup$ Excellent. Did you find $7$ by trial and error? $\endgroup$ – saulspatz Apr 5 '18 at 14:36
  • $\begingroup$ Basically. I was thinking about non-generators for powers of 2 so I could possibly get a contradiction, but I got something strong enough regardless. Originally, I was trying powers of 2, but there always seemed to be some value for which the LHS would have a large power of 2 as a factor. $\endgroup$ – Sharky Kesa Apr 5 '18 at 14:42
  • $\begingroup$ I tried 3 and 5 and didn't get anything interesting, so I quit. Should have kept going. $\endgroup$ – saulspatz Apr 5 '18 at 14:46
  • $\begingroup$ I didn't bother trying 3 or 5 because I knew they were generators for powers of 2. $\endgroup$ – Sharky Kesa Apr 5 '18 at 14:50

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