If $X = A^B + C$, why are C and X always divisible by A the same number of times?

Question

So I was playing around with numbers in my free time and I stumbled across this neat little trick,
If
$X = A^B + C$
Where
$A$ is any integer
$B$ is an integer greater than $1$
$C$ is not a multiple of $A^B$
Then $X$ and $C$ will both be divisible by $A$ the same number of times.
So for example
$3^3 - 6 = 21$, $6$ and $21$ are only divisible by $3$ once
$2^5 + 8 = 40$, $8$ and $40$ are divisible by $8$

I find this really interesting and was wondering why it works.

Answer 1

This can be simply explained through algebra and is a property of equations using integers. If you have an equation of the form

$$a=m(b+c)\,\,\,,\,\,\,a,b,c,m\in\mathbb Z\,\,\,,\,\,\,m\neq 0$$

then you can deduce that $a$ must be divisible by $m$ (as any property that applies to one side also applies to the other side of the equation as they're equal).

That's exactly what's happening in the example you give.

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First consider the case where $X=A^ix$ where $i\in\mathbb N_0$ and $x$ is not a multiple of $A$, then you can rewrite your equation as

$$X=A^ix=A^B+C$$ $\implies$ $$C=A^ix-A^B$$

If $i\ge B$, then we can conclude

$$C=A^B(A^{i-B}x-1)$$

which means $A^B$ divides $C$, breaking one of your conditions.

Therefore, $i<B$ and we get

$$C=A^i(x-A^{B-i})$$

which means $A^i$ divides $C$.

Now let us consider the case where $C=A^jc$ where $j\in\mathbb N_0$ and $C$ is not a multiple of $A$. Now our equation is

$$X=A^B+A^jc$$

If $j\ge B$ then we can again conclude

$$X=A^B(1-A^{j-B}c)$$

which means $A^B$ divides $C$ breaking one of your conditions.

Therefore, $j<B$ and we get

$$X=A^j(A^{B-j}+c)$$

and so $A^j$ divides $X$.

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Now we have shown:

If $A$ divides $X$ exactly $i$ times, then $A$ divides $C$ at least $i$ times.

If $A$ divides $C$ exactly $j$ times, then $A$ divides $X$ at least $j$ times.

Therefore, $A$ divides $X$ and $C$ the same number of times, leading to your conclusion.

Answer 2

Let $C= k \cdot A^n$ where $n$ is the biggest integer making $k$ an integer. Clearly, $k$ is not a multiple of $A$, and $C$ is divisible by $A$ only $n$ times.

Since $C$ is not a multiple of $A^B$, we must have $n < B$.

Then, $A^B+C = A^B + k \cdot A^n = A^n \big(A^{B-n} +k\big) =X$.

Observe, that $A^{B-n} +k$ is not a multiple of $A$, because $k$ is not a multiple of $A$.

Therefore $X$ also is divisible by $A$ only $n$ times.

Answer 3

Yes, this is true.

For proving this, let us suppose - $X$ is divisible by $A$, $m$ times $\implies X=p \cdot A^m$, and $C$, $n$ times.$\implies C=q \cdot A^n$

Now let us assume, $B>m,n$

Our equation is -

$$X=A^B+C$$

$$\implies p \cdot A^m=A^B+q \cdot A^n$$

Now if $m> n$, then - $$p \cdot A^m-A^B=q \cdot A^n$$

Here LHS is divisible by $A$ more than $m$ times, on the other hand RHS is divisible only $n$ times, giving us a contradiction. Similarly, the case $m<n$ will be rejected. Hence, the only possibility is $\color{blue}{m=n}3.

Similarly, this result can be proved even if $B<m$ and/or $B<n$.