Question

4. Prove that for positive integer n we have 169|3³ⁿ⁺³-26n-27. 

Answer 1

Let $P(n)$ be the proposition that $169|3^{3n+3}-26n-27$ for the positive integer $n.$

Basis Step

$P(1)$ is true because $3^{3(1)+3}-26(1)-27=676$ is divisible by $169.$

Inductive Step

Assume that $P(k)$ is true for an arbitrary nonnegative integer $k.$ That is, we assume that $3^{3k+3}-26k-27$ is divisible by $169.$

We must show that when we assume that $P(k)$ is true, then $P(k + 1)$ is also true. That is, we must show that $3^{3(k+1)+3}-26(k+1)-27$ is divisible by $169.$

We find that

$$3^{3(k+1)+3}-26(k+1)-27$$

$$=27(3^{3k+3})-26k-26-27$$

$$=27(3^{3k+3}-26k-27)+26(26k)+26(27)-26$$

$$=27(3^{3k+3}-26k-27)+676(k+1)$$

$676$ is divisible by $169,$and $3^{3k+3}-26k-27$ is divisible by $169$ by the inductive hypothesis. Then $27(3^{3k+3}-26k-27)+676(k+1)$ is divisible by $169.$ This completes the inductive step of the proof.

We have completed the basis step and the inductive step. Hence, by mathematical induction $P(n)$ is true for all positive integers $n.$

That is, we have proved that $169|3^{3n+3}-26n-27$ for the positive integer $n.$