Diophantine Equation with 2017th powers: $a^{2017}+a-2=(a-1)(b^{11})$ This problem stems from a recent student-created olympiad contest. 
Find all integer (not simply positive) solutions to $a^{2017}+a-2=(a-1)(b^{11})$. 
My multiple attempts modulo many small primes suggests that modulo arithmetic if it works will require an extremely contrived approach. I will note that (mod 23) did not work as expected. Furthermore note that $a=1$ is always a solution. (NOTE: There was an error in the problem statement that I didn't initially notice.)
 A: OK, so $a=1$ is one solution.  We wish to show there are no others, using the hint in your problem.
Hint: first, solve for $b^{11}$, in terms of $a$.  Changing my answer to account for the error you corrected, you would have:
$a^{2016}+\cdots+a^2+a^1+1+1 = b^{11}$
Second, reduce (mod 23).  Let $c=a^{-1}$.  You should get the equation 
$b^{11}=\frac{a^{2017}-1}{a-1} + 1= \frac{a^{-7}-1}{a-1} +1= \frac{c^7-1}{1/c-1}+1=-c(c^6+c^5+c^4+c^3+c^2+c+1)+1 =$
$= 1-(c^7+c^6+c^5+c^4+c^3+c^2+c)$
But, since $(b^{11})^2 = b^{22} \equiv b^0 = 1$ (mod 23), then $b^{11}$ is always either 1 or -1.
Then, just check all values of the LHS (mod 23) to see that this has no solution (mod 23).  They are, in order from $a=1$ to $a=23$:
[this calculation was apparently wrong]
None of these are congruent to 1 or -1 (mod 23), and thus you get that there are no solutions for $a$.
A: Here is some progress. I'm afraid it's probably not farther than you have already explored. 
Throughout, it is important to know that $11$th powers mod $23$ can only be $-1,0$, or $1$.
Of course, $a=1$ gives a solution. If $a\neq1$, it is OK to divide by $a-1$, giving you 
$$\begin{align}
\sum_{k=0}^{2016}a^k+1&=b^{11}\\
\end{align}$$
If $23\mid a$, then the equation reduces to $2\equiv b^{11}$, but that equation has no solution mod $23$.
If $23\not\mid a$, then 
$$\begin{align}
\sum_{k=2002}^{2016}a^k+\sum_{m=0}^{90}\sum_{n=0}^{21}a^{22m+n}+1&=b^{11}\\
\sum_{k=0}^{14}a^k+91\sum_{n=0}^{21}a^{n}+1&\equiv b^{11}\mod{23}\\
\sum_{k=0}^{14}a^k-\sum_{n=0}^{21}a^{n}+1&\equiv b^{11}\mod{23}\\
-{\sum_{n=15}^{21}a^{n}}+1&\equiv b^{11}\mod{23}\\
-a^{15}\sum_{n=0}^{6}a^{n}+1&\equiv b^{11}\mod{23}\\
\end{align}$$
If $a\equiv1$, the equation reduces to $-6\equiv b^{11}$, which again has no solution mod $23$. If $a\not\equiv1$, then
$$\begin{align}
-a^{15}\frac{a^7-1}{a-1}+1&\equiv b^{11}\mod{23}\\
\end{align}$$
It's a bit tedious, but we can evaluate the left side for $a\equiv2,3,\ldots22$. The left side never works out to be $1$ or $-1$. It can be $0$ though for $a\equiv11$ or $a\equiv15$. 
So $23\mid b$ and $a\equiv11$ or $a\equiv15$ mod $23$. I didn't find a way to leverage this into more help though. The best I could see from here is that even the smallest (in absolute value) option for $a$, which is $-8$, would imply an enormous (probably noninteger) value for $b$ with $166$ digits. This is somewhat clear from the original equation though, once you eliminate smaller options for $a$ directly.
