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If $a+d$, $(b-1)c$, $ab-a+c$ are divisible by $m$, prove that $ab^n+cn+d$ is also divisible by $m$.

I want to prove this by induction. For proving $ab^{k+1}+c(k+1)+d$ is divisible by $m$, i want to prove that $ab^k(b-1)+c$ is divisible by $m$ and then add it to $ab^{k}+ck+d$. Any idea how to prove $ab^k(b-1)+c$ is divisible by $m$? Or is there a better way to solve the problem?

Thanks in advance.

Ju Bc
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4 Answers4

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$\bmod m\!:\ \color{#0a0}{f_{\large n+1}\!-b f_{\large n}} =\, \overbrace{(1\!-\!b)c}^{\large \equiv\ 0}n \,\overbrace{-\color{#c00}d\,b+\color{#c00}d+c}^{\large \equiv\ 0\ \ {\rm by}\ \ \color{#c00}{d}\ \equiv\ -a}\!\equiv\color{#0a0} 0,\ $ so $\ f_n\equiv 0\,\Rightarrow\,\color{#0a0}{f_{n+1}\equiv bf_n\equiv} 0$

Bill Dubuque
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3

As I show here, such results boil down to the first $\,2\,$ terms of the Binomial Theorem, viz.

$\!\bmod m\!:\ c\equiv -a(b\!-\!1)\,$ so $\ \color{#0a0}0\equiv (b\!-\!1)c\equiv -\color{#0a0}{a(b\!-\!1)^2}$ so only $1$st $2$ terms survive below

$\qquad\quad\ \ \ a(1+b\!-\!1)^n =\, \color{#c00}{a\, +\,} n\,\underbrace{a(b\!-\!1)}_{\Large \equiv\ \color{#c00}{-c}}\ +\ \underbrace{\color{#0a0}{a(b\!-\!1)^2}(\cdots)}_{\large \equiv\ \color{#0a0}0}\ \,$ so adding $\ cn+d\ $ we get

$\quad\ \Rightarrow\ \ ab^n +cn+d\, \equiv\, \color{#c00}{a-c\,n}+cn\! +\! d \,\equiv\, a\!+\!d\,\equiv\, 0\ \ \ $ QED

Remark $ $ If you require an inductive proof then do so for the first $2$ terms in the Binomial Theorem. It's easy - the inductive step amounts to multiplying by $\,1\!+\!a\pmod{\!a^2},\,$ viz.

$\!\begin{align}{\rm mod}\,\ \color{#c00}{a^2}\!:\,\ (1+ a)^n\, \ \ \equiv&\,\ \ 1 + na\qquad\qquad\,\ \ {\rm i.e.}\ \ P(n)\\[1pt] \Rightarrow\ \ (1+a)^{\color{}{n+1}}\! \equiv &\ (1+na)(1 + a)\quad\, {\rm by}\ \ 1+a \ \ \rm times\ prior\\ \equiv &\,\ \ 1+ na+a+n\color{#c00}{a^2}\\ \equiv &\,\ \ 1\!+\! (n\!+\!1)a\qquad\ \ \ {\rm i.e.}\ \ P(\color{}{n\!+\!1})\\[2pt] \end{align}$

Bill Dubuque
  • 272,048
2

We need to proceed by

  • base case: $n=0 \implies ab^0+c0+d=a+d$
  • induction step: assume $m|ab^n+cn+d$ then

$$ab^{n+1}+c(n+1)+d=ab^{n+1}+bcn+bd-bcn-bd+c(n+1)+d=$$

$$=b(ab^{n}+cn+b)-(b-1)cn-bd+c+d$$

then note

  • $m|a+d \implies m|ab+bd \implies m|bd+a-c\implies m|bd-c-d$
user
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  • Well done. You might also enjoy the alternative method using the Binomial Theorem in my 2nd answer. It shows where the proposition comes from, i.e. it shows how to discover it, vs. how to verify it. – Bill Dubuque Oct 17 '18 at 00:52
  • @BillDubuque Thanks, I'll take a look to your solutions too. Bye – user Oct 17 '18 at 09:08
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Without induction. Let's do some tagging 1st $$m \mid a+d \tag{1}$$ $$m \mid (b-1)c \tag{2}$$ $$m \mid ab-a+c \tag{3}$$ then $$m \mid ab^n+cn+d \iff m \mid a\left(b^n-1\right)+cn+\color{red}{a+d} \overset{(1)}{\iff}\\ m \mid a\left(b^n-1\right)+cn \iff \\ m \mid a(b-1)\left(b^{n-1}+b^{n-2}+...+b^2+b+1\right)+cn \iff \\ m \mid (ab-a+c-c)\left(b^{n-1}+b^{n-2}+...+b^2+b+1\right)+cn \iff \\ m \mid \color{red}{(ab-a+c)\left(b^{n-1}+b^{n-2}+...+b^2+b+1\right)}-c\left(b^{n-1}+b^{n-2}+...+b^2+b+1\right)+cn \overset{(3)}{\iff}\\ m \mid c\left(b^{n-1}+b^{n-2}+...+b^2+b+1\right)-cn \iff \\ m \mid c\left(b^{n-1}-1\right)+c\left(b^{n-2}-1\right)+...+c\left(b^2-1\right)+\color{red}{c\left(b-1\right)} \overset{(2)}{\iff}\\ m \mid c\left(b^{n-1}-1\right)+c\left(b^{n-2}-1\right)+...+c\left(b^2-1\right)$$ which is true because for $\forall k\geq 2$ we have $$c\left(b^k-1\right)=\color{red}{c\left(b-1\right)}\left(b^{k-1}+b^{k-2}+...+b^{2}+b+1\right)$$ and from $(2)$ $$m \mid c\left(b^k-1\right)$$

rtybase
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