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For example $8$ is in the middle of the interval between $5$ and $11$, $9$ is at equal distance between $7$ and $11$; $10$ between $7$ and $13$.

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If so, then every even number is a sum of two primes. But this is a notorious open problem, known as the Goldbach conjecture.

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    For those to whom it's not obvious why: $n$ is exactly between $p$ and $q$ is equivalent to $2n = p+q$. And to be precise, the conjecture in the question is only equivalent to the conjecture that every even number *that is not twice a prime* is a sum of two primes, which is, strictly speaking, weaker than the Goldbach conjecture (though of course there's no reason to expect it to be easier or the answer to be different).2011-11-16
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    Also, it may be worth adding that the Goldbach conjecture is believed to be true and has been verified up to very large numbers, which means (for the OP) that (it is believed that) *every* number is at equal distance between two prime numbers.2011-11-16
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    @ShreevatsaR: It's not weaker than the Goldbach conjecture -- numbers that are twice a prime are trivially the sum of two primes...2011-11-16
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    Could we ask then: "Is every positive number larger then 1 at equal distance between two prime numbers?"2011-11-16
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    @VladDimitriu do you mean every positive non-prime number? If primes are allowed, surely 3 is a counterexample.2011-11-16
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    @kojiro I just read the explanation for 1 is not a prime number.2011-11-16
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    [2,5] is 3 spaces apart.2011-11-17
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    @Charles: Ah, and every prime is at equal distance between two primes. :-) Thanks.2011-11-17
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    @ShreevatsaR that can't be right. I was just trying to say above, 3 is not equidistant from 2 and 5.2011-11-17
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    @kojiro, 3 is equidistant from 3 and 3.2011-11-17
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    @GerryMyerson 3 is only one prime.2011-11-18
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    @kojiro, the question called for two prime numbers. It didn't call for two *distinct* prime numbers. Similarly, 6 is not a counterexample to "every even number exceeding 2 is a sum of two primes," since $6=3+3$ is a sum of two primes. It's just a convention, and you're welcome to use your own convention instead, but watch out, because everybody else will be using the convention you aren't using.2011-11-18
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    @GerryMyerson that happens to me all the time. But to be sure, I think I'm right to infer that the numbers should be [low, high] relative to the original nonprime, lest the proof become trivial and meaningless: 4 is equidistant from 3 and 3; 6 from 5 and 5; 8 from 7 and 7, etc… "Every nonprime int is equidistant from a prime and that same prime."2011-11-18
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    @kojiro, you have a point. Perhaps one needs to say for every $n\ge2$ there are primes $p,q$ with $q-n=n-p$. Anyway, for the Goldbach conjecture, one generally allows the two primes to be equal.2011-11-19
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    @koijro, I think that the word *between* in the question excludes saying that say 8 is at equal distance between 17 and 17. However one could argue that 3 is really not between 3 and 3 either (again it depends on language conventions).2011-11-24
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1 is a positive nonprime number not between any prime numbers at all. If you consider that cheating (I wouldn't know why), then see Gerry Myerson's anwer.

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    Surely 1 is a prime number? It's only divisible by 1 and itself (1)2011-11-16
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    1 isn't considered prime because, otherwise, various theorems would be inelagently phrased ("all primes except 1").2011-11-16
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    @LordScree: No, $1$ is not prime, it is a unit. You could see http://math.stackexchange.com/questions/120/is-1-a-prime-number2011-11-16
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    OK, thanks for clarifying for me :)2011-11-16
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    1used to be considered prime; indeed 1 is a necessary factor in "perfect" numbers2012-07-28
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    @AAA Consider posing questions or providing suitable answers to posted questions in order to gain the [50 reputation points](http://math.stackexchange.com/privileges/comment) required to begin commenting. (And note that to be [_perfect_](http://en.wikipedia.org/wiki/Perfect_number) a positive integer must equal the sum of _all_ its proper positive factors, not just the prime factors; _e.g._, $28 = 1 + 2 + 4 + 7 + 14$ is perfect.)2012-07-28
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    @ArthurFischer: I suppose you were addressing user AAA, but this is not completely clear, since "@AAA" is missing (curious, now that I posted this comment it becomes visible! so forget the current comment).2012-07-28
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    @MarcvanLeeuwen: Originally my comment was to a answer posted by AAA, and somehow that answer was later merged here (I assume sometime before you posted your comment to me). I guess them moderators were moderating (and my hat is off to them!)! Cheers!2012-07-28
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Check out a related theory: 'Green-Tao Theorem' which is a special case of Erdős conjecture and 'Primes in arithmetic progression' - in short, the primes contain arbitrarily long arithmetic progressions.

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    Copy + paste error, thanks for noticing my mistake. Re-edit time.2011-11-17
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Every prime number $>3$ also...

Every integer greater than 3 can be expressed as the average of two primes.

If a number is the average (or difference) of two primes, by doubling the number it has a partition of those two primes. So, for example, $(7+31)/2=19$ becomes $7+31=2∗19$. The Goldbach conjecture applies to even numbers only, but the average of two primes applies to every number - even, odd, prime - bigger than 3.

CSV of first 100,000: int, diff, p1, p2, type