proof by induction 5^n-1 is divisible by 4

**Jskid** · November 28th 2011, 06:25 PM

Use mathematical induction to prove that the assertion is true for $n\ge 1$ . $5^n-1$ is divisible by $4$ .

basis $n_{o}=1$ then $5^1-1=4$ is divisible by $4$ .

IH let $1\le l<k$ and assume $5^l-1$ is divisible by $4$ .

Induction Step
$5^{k}-1=(5)5^{k-1}-1$
Since $5^{k-1}-1$ is divisble by $4$ (by the induction hypothesis) then $(5)5^{k-1}-1$ is also divisible by $4$

Is this right?
What is wrong with the latex

**pickslides** · November 28th 2011, 06:32 PM

The induction step should have n = k+1

**Annatala** · November 29th 2011, 09:54 AM

IH let $1\le l<k$ and assume $5^l-1$ is divisible by $4$ .

If this is a proof by induction you need to go a step at a time, not prove it for all numbers at once.

I'd say this. "Assume $4 |(5^k-1)$ . We want to show $4|( 5^k^+^1-1)$ ."

Since $5^{k-1}-1$ is divisble by $4$ (by the induction hypothesis) then $(5)5^{k-1}-1$ is also divisible by $4$

Not quite, no. You have the formula wrong, and you need to show your work; how do you know 4 divides $5^k^+^1-1$ for certain?

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