is this enough prove?

**Petrus** · September 20th 2012, 09:41 AM

hey I should perhaps have seminar presentation on this task and wonder if that evidence is good enough?

show that the 5^(n +1) -1 is divisible by 4, for all n> _ = 0 (greater or equal to 0)

my solve:
5^(n+1)-1
5≡1 (mod 4)
5^(n+1) = 5^n*5≡ 1^n*1 (mod 4)
1^n*1-1≡0 (mod 4)
1^n = 1 n> _ = 0 (greater or equal to 0)
1-1≡0 (mod 4)
what i would prove

**Soroban** · September 20th 2012, 11:12 AM

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 $n\!+\!1$ power.

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

Are you trying an Inductive Proof?

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

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

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

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

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

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

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

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

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

We have proved $S(k+1).$
The inductive proof is complete.

**Petrus** · September 20th 2012, 11:25 AM

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?

**richard1234** · September 20th 2012, 11:34 AM

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

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

**Petrus** · September 20th 2012, 11:59 AM

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

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