Integer solutions of a variable coefficient polynomial I have many equations to solve similar to this one:
$$2 a b^3 - a b^2 + a b - 2 a - b^4 + b^3 - 2 b^2 + 2 b = 0$$
Here, b is a base and a is a non-zero digit in a b-adic number, so $1 \leq a \leq b-1$
I am looking for integer solutions to this equation. Short of writing a program to test loads of numbers, what are my options for solving this?

My attempts:
Taking the equation $\pmod b$ we get:
$$- 2 a \equiv 0 \pmod b$$
Which implies a = b/2
Taking the equation $\pmod a$ we get:
$$- b^4 +b^3 - 2b^2 + 2b \equiv 0 \pmod a$$
$$ b^3 + 2b \equiv b^4 + 2b^2 \pmod a$$
$$ b^3 + 2b \equiv b(b^3 + 2b) \pmod a$$
$$ 1 \equiv b \pmod a$$
^something seems wrong with that equation
The only solution I know of for this particular equation is:
$(b,a) = (2,1)$

Another equation that I have is this one:
$$-b^4 + (2a_2-a_1)b^3 + (2a_1+a_2+3)b^2 - (2a_1+a_2)b + a_1-2-2a_2 = 0$$
For which I computationally determined the solutions:
$(b,a_1,a_2) = (5,3,3)$
$(b,a_1,a_2) = (10,6,7)$
$(b,a_1,a_2) = (15,9,11)$
$(b,a_1,a_2) = (20,12,15)$
Which yielded the linear relation $(b,a_1,a_2) = (b, \frac{3b}{5}, \frac{4b-5}{5})$
Plugging $a_1 = \frac{3b}{5}, a_2 = \frac{4b-5}{5}$ into the equation works.
My motivation for this: The equations I have found link to fixed points in the Kaprekar routine of a base b number.
Thanks in advance for any and all help, Ben

Say $b   \ne 1$. Write $$ a (2b^3 -  b^2 +  b - 2 ) = b^4 - b^3 + 2 b^2 - 2 b $$
so $$2b^3 -  b^2 +  b - 2 \mid  b^4 - b^3 + 2 b^2 - 2 b $$
so $$2b^2+b+2\mid b(b^2+2)$$ 
If $b$ is odd then $\gcd(b,2b^2+b+2)=1$ so $$2b^2+b+2\mid b^2+2 \implies 2b^2+b+2\leq b^2+2 $$
which is impossible. 
So $b$ is even $b=2c$ and now $$4c^2+c+1\mid c(4c^2+2)$$ 
Since $\gcd(c,4c^2+c+1)=1$ we have $$4c^2+c+1\mid 4c^2+2\implies4c^2+c+1\leq 4c^2+2$$  so $c\leq 1$ and thus $c=1$, $b=2$ and $a=1$.
