I have this problem that I know ( by just knowing ) that its true, but I have to prove by induction, and I am stuck on the induction step.

The question asks:

Use induction to show that 5^n - 1 (5 to the power of n ) is divisible by 4 for all n element of Natural numbers , n >= 1.

So I have come this far:

Let P(n) be the predicate 4 | 5^n - 1

Then P(1) is 4|5^1 - 1 = 4 which is a tautology

We must show that for k>=1 if P(k) is true, then P(k+1) must also be true.

We assume for some k >=1 that

4| 5^n - 1

We now want to show that P(k+1) is true:

4 |5^(k+1) - 1

The right hand side of this statement is

5^(k+1) - 1 = 5^k.5 - 1

Now I know that it will work for all exponents of 5 because every exponential of 5 ends in 25 ( 125, 625, 3125, 15625) and minusing 1 gives 24 which is divisble by 4 ( and any multiple of 100 is also divisble by 4)

So how can i put this in induction language?

I have also tried it this way:

Let's say

5k - 1 = 4r where r is a natural number. We can rewrite this as :

5^k = 4r + 1

Basic step:

So for n=1 P(1) = 5-1=4(1) = 4,this is true.

Induction step:

Then for k+1:

5k+1 - 1 = 4r

5.5k = 4r + 1

This is the same as above (in bold) showing that for k+1 5k-1 is always divisible by 4.

Many thanks

The question asks:

Use induction to show that 5^n - 1 (5 to the power of n ) is divisible by 4 for all n element of Natural numbers , n >= 1.

So I have come this far:

**Basic Step:**Let P(n) be the predicate 4 | 5^n - 1

Then P(1) is 4|5^1 - 1 = 4 which is a tautology

**Induction Step:**We must show that for k>=1 if P(k) is true, then P(k+1) must also be true.

We assume for some k >=1 that

4| 5^n - 1

We now want to show that P(k+1) is true:

4 |5^(k+1) - 1

The right hand side of this statement is

5^(k+1) - 1 = 5^k.5 - 1

Now I know that it will work for all exponents of 5 because every exponential of 5 ends in 25 ( 125, 625, 3125, 15625) and minusing 1 gives 24 which is divisble by 4 ( and any multiple of 100 is also divisble by 4)

So how can i put this in induction language?

I have also tried it this way:

Let's say

5k - 1 = 4r where r is a natural number. We can rewrite this as :

5^k = 4r + 1

Basic step:

So for n=1 P(1) = 5-1=4(1) = 4,this is true.

Induction step:

Then for k+1:

5k+1 - 1 = 4r

5.5k = 4r + 1

This is the same as above (in bold) showing that for k+1 5k-1 is always divisible by 4.

Many thanks

Last edited: