Prove by method of induction, for all n ∈ N:

(2^{4n}−1) is divisible by 15

#### Solution

(2^{4n}−1) is divisible by 15 if and only if (2^{4n}−1) is a multiple of 15.

Let P(n) ≡ (2^{4n}−1) = 15m, where m ∈ N

**Step I:**

Put n = 1

∴ 2^{4(1)} − 1 = 16 − 1 = 15

∴ (2^{4n} − 1) is a multiple of 15.

∴ P(1) is true

**Step II:**

Let us consider that P(n) is true for n = k

i.e., 2^{4k} − 1 is a multiple of 15.

∴ 2^{4k} − 1 = 15a, where a ∈ N

∴ 2^{4k} = 15a + 1 ...(i)

**Step III:**

We have to prove that P(k + 1) is true

i.e., to prove that `2^(4("k"+1)) − 1` is a multiple of 15.

i.e., `2^(4("k"+1)) − 1` = 15b, where b ∈ N

∴ P(k + 1) = `2^(4("k"+1)) − 1 = 2^(4"k"+4) − 1`

= 2^{4k}.2^{4} − 1

= 16.(2^{4k}) − 1

= 16(15a + 1) − 1 ...[From (i)]

= 240a + 16 − 1

= 240a + 15

= 15(16a + 1)

= 15b, where b = (16a + 1) ∈ N

∴ P(k + 1) is true

**Step IV:**

From all steps above by the principle of mathematical induction, P(n) is true for all n ∈ N.

∴ (2^{4n} − 1) is divisible by 15, for all n ∈ N.