Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have thought a lot but am failing to arrive at anything encouraging.

First try: If this is to be proved by contradiction, then I start with the assumption that let $n$ be a number which is a sum of two numbers, of which at least one is prime. This gives $n = p + c$, where $p$ is the prime number and $c$ is the composite number. Also, any composite number can be written as a product of primes. So I can say, $n = p + p_1^{e_1}.p_2^{e_2}...p_k^{e_k}$. From this, I get $n - p = p_1^{e_1}.p_2^{e_2}...p_k^{e_k}$, but I have no clue what to do next.

Second try: For an instant let me forget about contradiction. Since $n > 11$, I can say that $n \geq 12$. This means that either $p \geq 6$ or $c \geq 6$. Again I'm not sure what to do next.

Finally, consider that the number 20 can be expressed in three different ways: $17+3$ (both prime), $16+4$ (both composite), and $18+2$ (one prime and one composite). This makes me wonder what we are trying to prove.

The textbook contains a hint, "Can all three of $n-4$, $n-6$, $n-8$ be prime?", but I'm sure what's so special about $4, 6, 8$ here.

share|improve this question
2  
At least one of the three numbers $n-4$, $n-6$, $n-8$ is divisible by a certain prime... –  anon Jul 24 '13 at 9:08
1  
(what we are trying to prove is that it exists at least one way to write a number greater than 11 as the sum of two composite numbers. You may partition it in many different ways: what matters is, at least one partition uses two composite numbers) –  mau Jul 24 '13 at 10:15

2 Answers 2

up vote 8 down vote accepted

Spoiler #1

You can write $n = (n - \varepsilon) + \varepsilon$, where $\varepsilon \in \{4, 6, 8\}$.

Spoiler #2

$n - \varepsilon > 3$, as $n > 11$.

Spoiler #3

One of the three numbers $n - \varepsilon$ is divisible by $3$, as they are distinct modulo $3$.

share|improve this answer
1  
Spoiler #4 >! nice spoiler(+1) –  Sami Ben Romdhane Jul 24 '13 at 9:07
    
@SamiBenRomdhane, thanks! –  Andreas Caranti Jul 24 '13 at 9:07
    
This is great! But where does this involve proof by contradiction? –  dotslash Jul 24 '13 at 9:26
    
It doesn't. So what? Why do you care how it's proved? –  Gerry Myerson Jul 24 '13 at 9:36
    
@Gerry Presumably because that's what's written in the textbook OP mentions (why it would say it wants a proof by contradiction given the nature of its hint I don't understand). –  anon Jul 24 '13 at 10:51

Only 9 even numbers greater than 4 can't be expressed as the ORDERED sum of two ODD composites, namely 6, 8, 10, 12, 14, 16, 22, 32, 38.

Look at the 4 identities: 1. pp(2n)=pr[2,n]-pc(2n) 2. cc(2n)=c[2,n]-cp(2n) 3. pp(2n)=pr[n,2n-2]-cp(2n) 4. cc(2n)=c[n,2n-2]-pc(2n)

where pp(2n)=number of ordered sum of 2 primes = 2n, cc(2n)=# of ordered sums of 2 composites=2n, cp(2n)=number of ordered sums of 1 composite and 1 prime (in that order)=2n, and pc(2n)= number of ordered sums of 1 prime and 1 composite (in that order)=2n, and a+b is an ordered sum iff a< or = to b, pr[a,b] = number of primes in[a,b], c[a,b] = number of composites in [a,b]

Lots of other identities to construct from the 4 above - have fun playing with.

share|improve this answer
    
and: pr[a,b] = the number of primes in [a,b] and c[a,b]= the number of composites in [a,b] –  d williams Dec 10 at 23:48
1  
For some basic information about writing math at this site see e.g. here, here, here and here. –  Chantry Cargill Dec 10 at 23:49

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.