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Prove 2^(3n) - 1 is divisible by 7 by induction

Step by step, with the reason for every step · Induction · Discrete Math

Step by step

The question

Question: prove by induction that 2³ⁿ − 1 is divisible by 7, for every whole number n.

23n − 1 is divisible by 7

Like dominoes: show it for n = 1, then show that k passes it on to k + 1.

1 Check n = 1

For n = 1: 7, and that is 7 × 1.

Divisible by 7: the first domino falls.

23 × 1 − 1 = 7 = 7 × 1

2 Assume it works for k

Suppose it works for k: 2³ᵏ − 1 is a multiple of 7.

So it is 7 times some whole number.

Call it m.

23k − 1 = 7m
m is a whole number

Move the rest to the other side: 2³ᵏ equals 7m + 1.

We will need this.

23k = 7m + 1

3 What we must show

We must show the same for k + 1: every n becomes k + 1.

= 23k + 3 − 1 is divisible by 7

Careful!

One more power means one more times the number.

It is 8 times as big, not 1 more.

Careful: not 23k + 3 = 23k + 1 ✗

4 Use the assumption

One more power: 2³ times the power with k.

23k + 3 − 1 = 8 × 23k − 1

Now swap in 7m + 1 for 2³ᵏ.

= 8(7m + 1) − 1

Open the bracket.

= 56m + 8 − 1

Like terms together.

= 56m + 7

5 Take out 7

Every term has a 7 in it.

Take out 7: 7(8m + 1).

= 7(8m + 1)
8m + 1 is a whole number

So it works for k + 1 too.

True for n = 1, and passed on each time: true for every whole number n.

The proof is done.

divisible by 7 for n = 1, and k → k + 1: for every n. ∎

6 Check

Let's check n = 3: it gives 511.

n = 3: 23 × 3 − 1 = 511

511 is 7 times 73.

Divisible by 7.

Right!

511 = 7 × 73

Let's look back.

First: check n = 1.

Then: assume it works for k.

Then: what we must show.

Then: use the assumption.

Then: take out 7.

Last: check.

Answer
8m + 1 is a whole number

Your turn

Try this one yourself, then check:

5n − 1 is divisible by 4
Show the answer
4(5m + 1)

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