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In how many ways can $10001$ be written as the sum of two primes?

Obviously since the 10001 is odd, one of the primes must be $2$. This leaves the second, must be prime as 9999, but it isn't, hence there are $0$ ways to write $10001$ as the sum of two primes.

Another way I would appreciate if someone can give me feedback on is:

All primes $p$ can be expressed as either $p=6k+1$ or $p=6k-1$, where $k$ is some positive integer. If I sum two distinct primes I have three possibilities: $$p_1+p_2=(6k+1)+(6m+1)=6(k+m)+2$$ or $$p_1+p_2=(6k-1)+(6m+1)=6(k+m)$$ or $$p_1+p_2=(6k-1)+(6m-1)=6(k+m)-2$$

Since $10001$ is neither even, which would satifie (1) or (3) and since it is not multiple of 6, which would satisfy (2), there exists no primes that would sum to $10001$. Is this logic correct?

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The first approach is good. For the second one, note that $2$ and $3$ are primes but they cannot be expressed as $6k\pm 1$ for a positive integer $k$. That may be a gap you may want to fill in. –  Lord Soth Oct 30 '13 at 2:22
    
Other than those small counter exceptions, the logic seems very good! –  Bennett Gardiner Oct 30 '13 at 2:23
    
That statement seems to be only valid for all primes 5 and above. I won't see how 3 falls into 6k+1 or 6k-1. Moreover, all three options results in an even answer. 103 can be written as the sum of 2 primes, 2+101. But 103 falls in neither category. –  imranfat Oct 30 '13 at 2:25
    
All primes EXCEPT $2$ and $3$ are of the form $6k\pm 1$. –  Michael Hardy Oct 30 '13 at 2:27

1 Answer 1

The primes 2 and 3 can't be expressed as 6k+1 or 6k-1 with k an integer, so you need to check those, but otherwise your solution is fine.

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