Re: is this enough prove?

Hello, Petrus!

It's hard to follow what you're doing.

. . I hope you don't present it like that . . .

And I don't like that $\displaystyle n\!+\!1$ power.

Quote:

$\displaystyle \text{Show that }5^n -1\text{ is divisible by 4, for all }n \ge 1$

Are you trying an Inductive Proof?

Verify $\displaystyle S(1)\!:\:5^1 - 1 \:=\:4$ . . . True!

Assume $\displaystyle S(k)\!:\;5^k - 1\text{ is divisible by 4.}$

That is: .$\displaystyle 5^k - 1 \:=\:4a\text{ for some intger }a.$

Add $\displaystyle 4\!\cdot\!5^k$ to both sides:

. . $\displaystyle 5^k - 1 + 4\!\cdot\!5^k \;=\;4a + 4\!\cdot\!5^k$

. . $\displaystyle 5^k + 4\!\cdot\!5^k - 1 \;=\;4(a + 5^k)$

. . $\displaystyle 5^k(4 + 1)-1 \;=\;4(a + 5^k)$

. . . . . $\displaystyle 5^k\!\cdot\!5 - 1 \;=\;4(a + 5^k)$

. . . . .$\displaystyle 5^{k+1} - 1 \;=\; \underbrace{4(a+5^k)}_{\text{divisible by 4}}$

We have proved $\displaystyle S(k+1).$

The inductive proof is complete.

Re: is this enough prove?

Hello!

That way u solved it is not how i would do but i got 2 question

1. The way i solved it would i get wrong on a test and i did solve it like that

2. I dont get it what happend with ^(n+1)?? On that i did 5^1-1 =4 do i mean that n=0?

Re: is this enough prove?

@Petrus, your proof seems correct. However you know that $\displaystyle 5 \equiv 1 (\mod 4)$ so therefore

$\displaystyle 5^{n+1} \equiv 1^{n+1} \equiv 1 (\mod 4)$ (this is a more direct method).

Re: is this enough prove?

Ohh never thought about it... Stupid of me sorry