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

[Math100]

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.

 

[Office_Shredder]

Don't use $n$ for $k-4$, you've already used it to mean $2k+1$!

 

[Math100]

Yes, you're right, I shouldn't use n, I'll just use $q$ then.

 

[Math100]

For replacement.

 

[Math100]

Other than that, is everything correct?

 

[fresh_42]

Other than that, is everything correct?

Yes. But "other than" is important. Never use the same variable name for two different variables.

 

[PeroK]

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

Where did you use the condition that $n > 11$?

 

[nuuskur]

Your idea works. For case #1 you may even consider $n-4$. The $n>11$ requirement isn't really effective here. Suffices that $n>6$.

For case #2, yes, it suffices to consider $n-9$ (because $9$ is the first odd composite on the list). One has $2k+1-9 = 2(k-4)$. If $n>11$, then surely $k>5$.

Note also that this statement is optimised. You can't relax the premise to $n\geqslant 11$, for example.