TITLE: Solve $2a^b-b=1997$ in $\mathbb{N}$
QUESTION [6 upvotes]: Problem. Find all pairs of positive integers $a$ and $b$ such that the following equality holds: $$2a^b-b=1997.$$

My attempt. We see that $b$ has to be odd, and that $L=(1997+b)/2$ has to be perfect power. Since $L\ge1998/2=999$, and since $L=2048$ gives $a=2$ and $b=11$ which is not solution, we can assume that $L\equiv1\pmod2$ (everything beyond $2^{13}$ is too big anyway). So $L$ is odd which means that $1997+b$ is not divisible by $4$ or: $$1997+b\not\equiv0\pmod4\iff b\not\equiv-1997\pmod4\iff b\not\equiv3\pmod4.$$
This gives us a few hits about $a^b$, but let's face it, that's pretty much useless. I'm clueless. I tried to nail the solution but it didn't work out, except the obvious one for $b=1$ which gives us $a=999$.

REPLY [7 votes]: Clearly there are no solutions with $a=1$, so we can assume that $a\ge2$. If $b\ge10$, then $2a^b\ge2\cdot2^{10}=2048$ which is too large (subtracting $b$ won't help, because $f(x)=2^x-x$ is increasing when $x\in[2,\infty)$). So $b=1,2,3,4,5,6,7,8$ or $9$.
Quick checking leaves $b=1, a=999$ and $b=3, a=10$ as the only possibilities.