# Diophantine Equation Concerning Powers of 2

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?

• Tip: You have to put $signs around the MathJax for it to get formatted. – saulspatz Apr 5 '18 at 1:45 • I'm going to guess that$m=2,n=3$is the only solution. This is just based on a quick computer experiment. – saulspatz Apr 5 '18 at 1:52 • I thought it might have to do with the lifting theorem but the best I've been able to do is prove that$m$must be even, which isn't a great deal of help. – saulspatz Apr 5 '18 at 4:33 • 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. – saulspatz Apr 5 '18 at 6:06 ## 1 Answer 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$(2,3)$. • Excellent. Did you find$7\$ by trial and error? – saulspatz Apr 5 '18 at 14:36
• 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. – Sharky Kesa Apr 5 '18 at 14:42
• I tried 3 and 5 and didn't get anything interesting, so I quit. Should have kept going. – saulspatz Apr 5 '18 at 14:46
• I didn't bother trying 3 or 5 because I knew they were generators for powers of 2. – Sharky Kesa Apr 5 '18 at 14:50