# [ASK] Mathematical Induction: Prove 7^n-2^n is divisible by 5.

#### Monoxdifly

##### Well-known member
Prove by mathematical induction that $$\displaystyle 7^n-2^n$$ is divisible by 5.

What I've done so far:

For n = 1

$$\displaystyle 7^1-2^1=7-2=5$$ (true that it is divisible by 5)

For n = k

$$\displaystyle 7^k-2^k=5a$$ (assumed to be true that it is divisible by 5)

For n = k + 1

$$\displaystyle 7^{k+1}-2^{k+1}=7^k\cdot7-2^k\cdot2=7(7^k-2^k)+12\cdot2^k=7(5a)+12\cdot2^k$$

This is where the problem lies. How can I show that $$\displaystyle 12\cdot2^k$$ is divisible by 5?

#### castor28

##### Well-known member
MHB Math Scholar
Prove by mathematical induction that $$\displaystyle 7^n-2^n$$ is divisible by 5.

What I've done so far:

For n = 1

$$\displaystyle 7^1-2^1=7-2=5$$ (true that it is divisible by 5)

For n = k

$$\displaystyle 7^k-2^k=5a$$ (assumed to be true that it is divisible by 5)

For n = k + 1

$$\displaystyle 7^{k+1}-2^{k+1}=7^k\cdot7-2^k\cdot2=7(7^k-2^k)+12\cdot2^k=7(5a)+12\cdot2^k$$

This is where the problem lies. How can I show that $$\displaystyle 12\cdot2^k$$ is divisible by 5?
Hi Monoxdifly ,

$12\cdot2^k$ is certainly not divisible by $5$ (look at the prime factors).

$$7^k\cdot7 - 2^k\cdot2 = 7(7^k-2^k) + 5\cdot2^k$$
and this should clear things up.

#### Monoxdifly

##### Well-known member
Ah, I see. Thank you very much!

#### Prove It

##### Well-known member
MHB Math Helper
It seems a bit overkill to use Induction, when there's a simple rule for the Difference of Two Terms With the Same Power...

\displaystyle \begin{align*} a^n - b^n \equiv \left( a - b \right) \sum_{r = 0}^{n - 1}{ a^{n - 1 - r}\,b^r } \end{align*}

so in your case the factor would be (7 - 2) which equals 5.

#### Monoxdifly

##### Well-known member
It seems a bit overkill to use Induction, when there's a simple rule for the Difference of Two Terms With the Same Power...

\displaystyle \begin{align*} a^n - b^n \equiv \left( a - b \right) \sum_{r = 0}^{n - 1}{ a^{n - 1 - r}\,b^r } \end{align*}

so in your case the factor would be (7 - 2) which equals 5.
Well, the school curriculum doesn't teach this rule, so yes, we were supposed to solve it with the "overkill" method.

#### Prove It

##### Well-known member
MHB Math Helper
Well, the school curriculum doesn't teach this rule, so yes, we were supposed to solve it with the "overkill" method.
Then you can consider it as something new that you have learnt. The most concise method of proof is always the best.