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?
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 $(2,3)$.