# Elementary Number Theory Problems 3.3 Solution (David M. Burton's 7th Edition) - Q9

My Solution for "(a) For $n > 3$, show that the integers $n$, $n + 2$, $n + 4$ cannot all be prime. (b) Three integers $p$, $p + 2$, $p + 6$, which are all prime, are called a $\textit{prime-triplet}$. Find five sets of prime-triplets."

## 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

(a) For $n > 3$, show that the integers $n$, $n + 2$, $n + 4$ cannot all be prime.
(b) Three integers $p$, $p + 2$, $p + 6$, which are all prime, are called a $\textit{prime-triplet}$. Find five sets of prime-triplets.

## Solution

### (a)

By Theorem 2.1 Division Algorithm, we can write $n$ in the form of $3k$, $3k + 1$, or $3k + 2$ for some integer $k$. If $n$ is prime, $n$ can either be $3k + 1$ or $3k + 2$.

If $n$ is $3k + 1$, then $n + 2 = 3k + 3 = 3(k + 1)$, which is divisible by $3$. The only prime will be $3$, which is impossible as $n > 3$.

If $n$ is $3k + 2$, then $n + 4 = 3k + 6 = 3(k + 2)$, which is also divisible by $3$. The only prime will be $3$, which is also impossible as $n > 3$.

Therefore, integers $n$, $n + 2$, $n + 4$ cannot all be prime.

### (b)

$5, 7, 11$
$11, 13, 17$
$17, 19, 23$
$41, 43, 47$
$101, 103, 107$

Read More: All My Solutions for This Book

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