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let $a,b,c,d$ be all positive integers such that $a-bc \neq 0$,and $\gcd(a,b)=1$. Under what conditions, $(a-bc)$|$(a-b^d)$? In other words, does it exist any integer $k \neq 1 $ such that:$$k=\dfrac{a-b^d}{a-bc}$$ I only can find the obvious solution $$ k=1$$ Are there any other ones?

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    $\begingroup$ this question needs a better title. $\endgroup$
    – abcd
    Commented Apr 8, 2015 at 17:15

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Let $a=4,$ $\quad$ $b=6,$ $\quad$ $c=1, \quad$ and $\quad$ $d=3$ $${{4-6^3} \over {4-6 \cdot 1}}=106$$ Its should start becoming apparent that isolated examples like this will crop up all day, so lets really get our hands dirty with this...

First off, lets get rid of the parts of this problem that constrain us, i.e. the division. Also let, $e=b^d$ and $f=b \cdot c$ $$k={{a-e} \over {a-f}}$$ $$k \cdot (a-f)=a-e$$ $$e=a-k \cdot (a-f)$$ Now you pick a k and the goal is to keep $e$ positive. I'll now show that this equation has an infinite number of solutions.

For $e$ to be positive... $$a \gt k \cdot (a-f)$$

$$a \gt ka-kf$$ $$a \cdot (1-k) \gt -kf$$ $$(k-1) \cdot a \lt kf$$ $$a \lt {{k} \over {k-1}} \cdot f$$ a is an integer, so in reality... $$a \lt floor \left({{k} \over {k-1}} \cdot f \right)$$ Demonstrably an infinite number of solutions, just pick a k. Enjoy!

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Let $a=73, b=3, c=19, d=2$

Thus $\frac{73-3^2}{73-3(19)} = \frac{64}{16} = 4$

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