# Elementary Number Theory Problems 4.2 Solution (David M. Burton's 7th Edition) - Q7

My Solution for "For $n \geq 1$, show that $$(-13)^{n + 1} \equiv (-13)^{n} + (-13) ^{n - 1} \pmod {181}$$ [Hint: Notice that $(-13)^{2} \equiv -13 + 1 \pmod {181}$; use induction on $n$.]"

## Background

All theorems, corollaries, and definitions listed in the book's order:

I will only use theorems or facts that are proved before this question. So you will not see that I quote theorems or facts from the later chapters.

## Question

For $n \geq 1$, show that $$(-13)^{n + 1} \equiv (-13)^{n} + (-13) ^{n - 1} \pmod {181}$$
[Hint: Notice that $(-13)^{2} \equiv -13 + 1 \pmod {181}$; use induction on $n$.]

## Solution

We can prove the statement by using mathematical induction.

Base Case:

For $n = 1$, $(-13)^{2} \equiv -13 + 1 \pmod {181}$.

Induction Hypothesis:

Assume $(-13)^{k + 1} \equiv (-13)^{k} + (-13) ^{k - 1} \pmod {181}$.

Consider the case for $k + 1$:

$$$$\begin{split} (-13)^{k + 2} & \equiv (-13)(-13)^{k + 1} \\ & \equiv (-13)[(-13)^{k} + (-13) ^{k - 1}] \quad \text{by Induction Hypothesis} \\ & \equiv (-13)^{k + 1} + (-13) ^{k} \pmod {181} \end{split} \nonumber$$$$

Conclusion:

This proves the statement by induction.

Read More: All My Solutions for This Book

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