Elementary Number Theory Problems 3.3 Solution (David M. Burton's 7th Edition) - Q23

Background

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

Theorems and Corollaries in Elementary Number Theory (Ch 1 - 3)

All theorems and corollaries mentioned in David M. Burton’s Elementary Number Theory are listed by following the book’s order. (7th Edition) (Currently Ch 1 - 3)

RanblogRan

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) The arithmetic mean of the twin primes $5$ and $7$ is the triangular number $6$. Are there any other twin primes with a triangular mean?
(b) The arithmetic mean of the twin primes $3$ and $5$ is the perfect square $4$. Are there any other twin primes with a square mean?

Solution

(a)

The definition of triangular number is $T_{n} = \frac{n(n + 1)}{2}$ for $n \in Z^{+}$. Let $p$ and $p + 2$ be the twin primes. We know:

From Theorem 3.2 Fundamental Theorem of Arithmetic, either $n + 2 = 2$ or $n - 1 = 2$. Then either $n = 0$ or $n = 3$. It is impossible for $n = 0$ as the right-hand side will become $-2$ while the left-hand side must be greater than $0$. So, $n = 3$. Therefore, $p = 5$, which is the only solution.

(b)

Similarly, we let $p$ and $p + 2$ be the twin primes and let $n^{2}$ be the perfect square. We then have:

Then, from Theorem 3.2 Fundamental Theorem of Arithmetic, either $n + 1 = 1$ or $n - 1 = 1$. It means that either $n = 0$ or $n = 2$. It is impossible for $n = 0$ by similar arguments as above. So, $n = 2$. Therefore, $p = 3$, which is the only solution.

Read More: All My Solutions for This Book

Related Pages

< Chapter 3.3, Q22 Chapter 3.3, Q24 >