- #1

Math100

- 780

- 220

- Homework Statement
- Establish the following statement:

Each integer n>11 can be written as the sum of two composite numbers.

[Hint: If n is even, say n=2k, then n-6=2(k-3); for n odd, consider the integer n-9.]

- Relevant Equations
- None.

Proof:

Suppose n is an integer such that ## n>11 ##.

Then n is either even or odd.

Now we consider these two cases separately.

Case #1: Let n be an even integer.

Then we have ## n=2k ## for some ## k\in\mathbb{Z} ##.

Consider the integer ## n-6 ##.

Note that ## n-6=2k-6 ##

=## 2(k-3) ##.

This means ## n=2(k-3)+6 ##

=## 2m+6 ##,

where ## m=k-3 ## is an integer.

Thus, ## 2m ## and ## 6 ## are two composite numbers.

Case #2: Let n be an odd integer.

Then we have ## n=2k+1 ## for some ## k\in\mathbb{z} ##.

Consider the integer ## n-9 ##.

Note that ## n-9=2k+1-9 ##

=## 2k-8 ##

=## 2(k-4) ##.

This means ## n=2(k-4)+9 ##

=## 2n+9 ##,

where ## n=k-4 ## is an integer.

Thus, ## 2n ## and ## 9 ## are two composite numbers.

Therefore, each integer ## n>11 ## can be written as the sum of two composite numbers.

Suppose n is an integer such that ## n>11 ##.

Then n is either even or odd.

Now we consider these two cases separately.

Case #1: Let n be an even integer.

Then we have ## n=2k ## for some ## k\in\mathbb{Z} ##.

Consider the integer ## n-6 ##.

Note that ## n-6=2k-6 ##

=## 2(k-3) ##.

This means ## n=2(k-3)+6 ##

=## 2m+6 ##,

where ## m=k-3 ## is an integer.

Thus, ## 2m ## and ## 6 ## are two composite numbers.

Case #2: Let n be an odd integer.

Then we have ## n=2k+1 ## for some ## k\in\mathbb{z} ##.

Consider the integer ## n-9 ##.

Note that ## n-9=2k+1-9 ##

=## 2k-8 ##

=## 2(k-4) ##.

This means ## n=2(k-4)+9 ##

=## 2n+9 ##,

where ## n=k-4 ## is an integer.

Thus, ## 2n ## and ## 9 ## are two composite numbers.

Therefore, each integer ## n>11 ## can be written as the sum of two composite numbers.