2023/016) Show that $6^{n+2}+7^{2n+1}$ for n positive integer is divisible by 43

 We shall prove the same by principle of mathematical induction.

Let $P(n) = 6^{n+2}+7^{2n+1}$ 

We shall show that base step and induction step are true

Base step

Here we show that P(1) is divisible by 43

$P(1) = 6^ 3 + 7^3 = 216 + 343 = 559 = 43 * 13$ is divisible by 43

So base step is true

Now for induction step

Induction step

Let it to true k that is P(k) is divisible of 43

We shall show that P(k+1) is divisible by 43

P(k) is divisible by 43 so there exists an integer b such that

$P(k) = 6^{k+2}+7^{2k+1}= 43b$ 

We have $P(k+1) = 6^{(k+1)+2}+7^{2(k+1)+1}$ 

$= 6(6^{k+1}) + 49 * 7^{2k+1} $

$= 6(6^{k+1}) + (6+43) * 7^{2k+1} $

$= 6(6^{k+1}) + 6 * 7^{2k+1} + 43 * 7^{2k+1}$

$= 6(6^{k+1} + 7^{2k+1}) + 43 * 7^{2k+1}$

$= 6 * 43 b + 43 * 7^{2k+1}$

$= 43( 6b + 7^{2k+1})$

This is multiple of 43

As we have proved both the base step and induction step hence this is true hence proved