Cfrgtkky

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

The first equation

$$2 a b^3 - a b^2 + a b - 2 a - b^4 + b^3 - 2 b^2 + 2 b = 0$$

can be written as

$$(b - 1) (2 a b^2 + a b + 2 a - b^3 - 2 b)=0$$

Since we have $bge 2$, dividing the both sides by $b-1 (not=0)$ gives

$$2 a b^2 + a b + 2 a - b^3 - 2 b=0$$
Solving for $a$ gives
$$a=frac{2b-1}{4}+frac{5b+2}{4(2b^2+b+2)}$$
Multiplying the both sides by $4$ gives
$$4a-2b+1=frac{5b+2}{2b^2+b+2}$$

Since the RHS has to be a positive integer, we have to have
$$frac{5b+2}{2b^2+b+2}ge 1,$$
i.e.
$$0le ble 2$$
from which $b=2$ follows.

So, the only solution for the first equation is $$color{red}{(a,b)=(1,2)}$$

The second equation
$$-b^4 + (2a_2-a_1)b^3 + (2a_1+a_2+3)b^2 - (2a_1+a_2)b + a_1-2-2a_2 = 0$$
can be written as
$$(1-b) (b^3 + b^2 a_1 - 2 b^2 a_2 + b^2 - b a_1 - 3 b a_2 - 2 b + a_1 - 2 a_2 - 2)=0$$

Since we have $bge 2$, dividing the both sides by $1-b (not=0)$ gives
$$b^3 + b^2 a_1 - 2 b^2 a_2 + b^2 - b a_1 - 3 b a_2 - 2 b + a_1 - 2 a_2 - 2=0$$
which can be written as
$$(b^2-b+1)a_1+{-2(b^2-b+1)-5b}a_2+{(b+2)(b^2-b+1)-b-4}=0$$

So, we have
$$5ba_2equiv -b-4pmod{b^2-b+1}$$
Multiplying the both sides by $b-1$ gives
$$5b(b-1)a_2equiv (b-1)(-b-4)pmod{b^2-b+1},$$
i.e.
$$5{(b^2-b+1)-1}a_2equiv -(b^2-b+1)-4b+5pmod{b^2-b+1},$$
i.e.
$$5a_2equiv 4b-5pmod{b^2-b+1}$$
So, there has to be an integer $k$ such that
$$5a_2=(b^2-b+1)k+4b-5$$
Using $1le a_2le b-1$, we have
$$5le (b^2-b+1)k+4b-5le 5b-5,$$
i.e.
$$frac{-4b+10}{b^2-b+1}le kle frac{b}{b^2-b+1}$$

Let $f(b),g(b)$ be the left fraction and the right fraction respectively.

Then, we have $f(2)=frac 23=g(2)$.

Also, we have $-1lt f(b)lt 0$ and $0lt g(b)lt 1$ for $bge 3$.

So, $k=0$ for $bge 3$ from which
$$5a_2=4b-5$$
follows.

It follows that $b=5m$ where $m$ is a positive integer and that $$a_2=4m-1,qquad a_1=3m$$ and these are sufficient.

Therefore, the only solutions for the second equation are
$$color{red}{(b,a_1,a_2)=(5m,3m,4m-1)}$$
where $m$ is any positive integer.

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 $$2b^2+b+2mid b(b^2+2)$$

If $b$ is odd then $gcd(b,2b^2+b+2)=1$ so $$2b^2+b+2mid b^2+2 implies 2b^2+b+2leq b^2+2 $$
which is impossible.

So $b$ is even $b=2c$ and now $$4c^2+c+1mid c(4c^2+2)$$

Since $gcd(c,4c^2+c+1)=1$ we have $$4c^2+c+1mid 4c^2+2implies4c^2+c+1leq 4c^2+2$$ so $cleq 1$ and thus $c=1$, $b=2$ and $a=1$.