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I have to prove the following statement using mathematical induction.

For all integers, $n \ge 1, 5^{2n} - 4^{2n}$ is divisible by 9.

I got the base case which is if $n = 1$ and when you plug it in to the equation above you get 9 and 9 is divisible by 9.

Now the inductive step is where I'm stuck.

I got the inductive hypothesis which is $ 5^{2k} - 4^{2k}$

Now if P(k) is true than P(k+1) must be true. $ 5^{2(k+1)} - 4^{2(k+1)}$

These are the step I gotten so far until I get stuck:

$$ 5^{2k+2} - 4^{2k+2} $$ $$ = 5^{2k}\cdot 5^{2} - 4^{2k} \cdot 4{^2} $$ $$ = 5^{2k}\cdot 25 - 4^{2k} \cdot 16 $$

Now after this I have no idea what to do. Any help is appreciated.

David K
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shawn edward
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  • Adding and subtracting the same thing is just like "adding zero" just in a tricky sort of way, and you are always allowed to add zero just as you are always allowed to multiply by one without changing anything. – JMoravitz Feb 16 '17 at 22:03
  • Main correction thus far is that the inductive hypothesis for $n=k$ is $ 9\mid(5^{2k} - 4^{2k})$, which can also be expressed by $(5^{2k} - 4^{2k})\equiv 0\pmod 9$, or, alternatively "Exits $m\in \mathbb Z$ such that $9m= 5^{2k} - 4^{2k}$" – amWhy Feb 16 '17 at 22:07

4 Answers4

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You're very close. Now add and subtract $4^{2k}$ in the first term to obtain $$ 5^{2k}\cdot 25-4^{2k}\cdot 16=25\cdot (5^{2k}-4^{2k})+(25-16)\cdot 4^{2k}=25\cdot (5^{2k}-4^{2k})+9\cdot 4^{2k} $$

The first term is divisible by $9$ by the induction hypothesis, hence the whole expression is divisible by $9$.

carmichael561
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  • I dont get how like you get two 25 after the first equal sign. – shawn edward Feb 16 '17 at 22:04
  • The $5^{2k}$ in the first expression becomes $5^{2k}-4^{2k}+4^{2k}$, and then I rearranged the terms. – carmichael561 Feb 16 '17 at 22:05
  • Thank you I understood it but these concept are weird, like knowing when to add and subtract a number and stuff like that. – shawn edward Feb 16 '17 at 22:14
  • Some of it comes from experience, but also since you know that $5^{2k}-4^{2k}$ is divisible by $9$ it stands to reason that you should try to make such a term appear. – carmichael561 Feb 16 '17 at 22:16
  • @shawn It appears weird because when the proof is presented this way it is completely devoid of arithmetical intuition. There are other ways to present the proof that highlight the arithmetical structure, which will become clearer when one learns modular arithmetic. I show one in may answer. I also highlight the key role of the Congruence Product Rule in many similar inductive problems, e.g. see the links here.. – Bill Dubuque Feb 16 '17 at 22:38
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The induction hypothesis can be written $$ 5^{2k}-4^{2k}=9m $$ for some integer $m$. Therefore $5^{2k}=4^{2k}+9m$ and so $$ 5^2\cdot5^{2k}-4^2\cdot4^{2k}= 25(9m+4^{2k})-16\cdot4^{2k}= 9\cdot 25m+(25-16)\cdot 4^{2k}= 9\cdot 25m-9\cdot 4^{2k} $$

Alternatively, $4\equiv -5\pmod{9}$, so $$ 5^{2k}-4^{2k}\equiv 5^{2k}-(-5)^{2k}\equiv 5^{2k}-5^{2k}\pmod{9} $$

egreg
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$\begin{align}{\bf Hint}\qquad\qquad\qquad\qquad\,\ \color{#c00}{25} &=\,\ \color{#c00}{16 + 9}\\ 25^{\large N} &=\,\ 16^{\large N}\! +\! 9j\\ \Rightarrow\,\ 25^{\large N+1}\! = \color{#c00}{25}\cdot 25^{\large N} &= (16^{\large N}\!+\!9j)(\color{#c00}{16+\!9}) = 16^{\large N+1} +9\,(\cdots)\ \end{align}$

Or, said mod $\,9\!:\,\ \begin{align} 25&\equiv 16\\ 25^{\large N}&\equiv 16^{\large N}\end{align}\ \Rightarrow\, 25^{\large N+1}\equiv 16^{\large N+1}\,$ by the Congruence Product Rule

Or, $ $ equivalently, $\ \big[25\equiv 16\big]^{\large N}\!\Rightarrow\, 25^{\large N}\!\equiv 16^{\large N}\, $ by the Congruence Power Rule, which is an inductive extension of the Product Rule.

Community
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Bill Dubuque
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The inductive hypothesis is

$$(5^{2n}-4^{2n})\bmod9=0.$$

Then

$$(5^{2n+2}-4^{2n})\bmod9=(25\cdot5^{2n}-16\cdot4^{2n})\bmod9=(9\cdot5^{2n}+16(5^{2n}-4^{2n}))\bmod9=0.$$