# How can I use induction solve this?

How can I show/solve this? I've tried by using the basis step and the inductive step, but just can't seem to get it right.

$$\forall(n \geq 0)(4\mid(9^n − 5^n)).$$

• do you mean $9^n-5^n$ by any chance? – lab bhattacharjee Apr 25 '13 at 14:58
• Yeah, sorry - overlooked that – Maren Apr 25 '13 at 14:59
• This is helpful. – Git Gud Apr 25 '13 at 15:07

You can use

$$9^n-5^n=(9-5)(9^{n-1}+9^{n-2}\cdot 5+..+5^{n-1})$$

This solution has hidden induction in it.

Alternately, by induction:

$$9^{n+1}-5^{n+1}=9^{n+1}-9^{n}\cdot 5+9^{n}5-5^{n+1}=9^n\cdot 4+5(9^n-5^n)$$

use $P(n)$ and done.

$(a^n-b^n)(a+b)=a^{n+1}-b^{n+1}+ab(a^{n-1}-b^{n-1})$

$\implies a^{n+1}-b^{n+1}=(a^n-b^n)(a+b)-ab(a^{n-1}-b^{n-1})$

If $a,b,n$ are positive integers,

if $(a-b)$ divides $(a^n-b^n)$ and $(a^{n-1}-b^{n-1}),$ it will divide $a^{n+1}-b^{n+1}$

Now for $n=0, a^0-b^0=0$ which is divisible by $(a-b)$ for $ab(a-b)\ne0$

for $n=1, a^1-b^1=a-b$

Alternatively, the way N.S. has solved it : $a^{n+1}-b^{n+1}=a^n(a-b)+b(a^n-b^n)$

• The latter is a special case of the proof of the Congruence Product Rule - see my answer. – Math Gems Apr 25 '13 at 20:15

$\begin{eqnarray}{\bf Hint}&&\rm m\mid\ \ \color{#C00}{a^k\!-\!b^k}\:\ and\:\ m\mid \color{#0A0}{a-b} \\ \Rightarrow\: &&\rm m\mid (\color{#C00}{a^k\!-b^k})\,a\,+\,b^k(\color{#0A0}{a-b})\, =\: a^{k+1}\!-b^{k+1}\end{eqnarray}$

Note $\$ The above proof is essentially a special case of that for the Congruence Product Rule. Below are proofs of this product rule proof expressed in both divisibility and congruence form,  using the standard notation: $\rm\ \ a\mid b \ :=\ a\,$ divides $\rm\, b,\$ and $\rm\ \, a\equiv b\ \ (mod\ m) \iff m\mid a-b$

$\begin{eqnarray} \rm {\bf Lemma}\ \ &\rm m\ \ |&\rm\ \, X\!-\!x\quad\ \& &&\rm\! m\ |\: Y\!-\!y \ \Rightarrow\ m\:|\!\!&&\rm XY - \: xy\\ \\ \rm {\bf Proof}\ \ \ \ \ &\rm m\ \ |&\rm (X\!-\!\color{#c00}x)\:\color{#c00}Y\ \,+ &&\rm\, \color{#c00}x\ (\color{#c00}Y\!-\!y)\ \ \ \ = &&\rm XY - \: xy \\ \\ \rm {\bf Lemma}\ \ & &\rm\ \, X\equiv x\quad\ \ \& &&\rm\quad\ Y\equiv y \ \ \ \ \Rightarrow\ &&\rm XY\equiv xy\\ \\ \rm {\bf Proof}\ \ \ \ \ &0\equiv& \rm (X\!-\!\color{#c00}x)\:\color{#c00}Y\,\ + &&\rm\, \color{#C00}x\ (\color{#c00}Y\!-\!y)\ \ \ \ \equiv &&\rm XY - \: xy \\ \end{eqnarray}$