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How to prove that each sum of odd prime and odd semiprime can be written as sum of two odd primes $(p_1+p_2p_3=p_4+p_5)$ ? Since we know that each prime number greater than $3$ is of the form $6k\pm 1$ we can take one of the combinations and write:

$p_1=6k_1+1 ; p_2=6k_2+1 ; p_3=6k_3+1 \Rightarrow p_1+p_2p_3=(6k_1+1)+(6k_2+1)(6k_3+1)=$

$=6k_1+1+6k_2+6k_3+36k_2k_3+1=6(k_1+k_2+k_3)+1+6(6k_2k_3)+1=$

$=(6k_4+1)+(6k_5+1)$ , Similarly we can show that for each combination of the $p_1 , p_2 ,p_3$ forms we get $(6k_4\pm 1)+(6k_5\pm 1)$ form. So we have necessary but not sufficient condition for $p_4$ and $p_5$ to be primes. Any idea how to find a sufficient condition for this proof ?

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In short, what you are stating should be true but it's not proven yet. By Chen's theorem, every (sufficiently large) even number is either a sum of two primes or a prime and semiprime. If we could show your question, then every (sufficiently large) even number is a sum of 2 primes. That is, we would have more-or-less proved the Goldbach conjecture... – Srivatsan Sep 24 '11 at 8:32
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@SrivatsanNarayanan,In my opinion this statement is weaker than Goldbach conjecture – pedja Sep 24 '11 at 8:35
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Strictly speaking, it is weaker because you will prove it only for sufficiently large even numbers. – Srivatsan Sep 24 '11 at 8:41
    
+1 great question – draks ... Mar 22 '12 at 22:27

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