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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 $13$.

Karatuğ Ozan Bircan
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Vlad Dimitriu
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4 Answers4

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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.

Gerry Myerson
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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). – ShreevatsaR Nov 16 '11 at 12:36
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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. – ShreevatsaR Nov 16 '11 at 12:57
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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... – Charles Nov 16 '11 at 14:22
  • Could we ask then: "Is every positive number larger then 1 at equal distance between two prime numbers?" – Vlad Dimitriu Nov 16 '11 at 14:37
  • @VladDimitriu do you mean every positive non-prime number? If primes are allowed, surely 3 is a counterexample. – kojiro Nov 16 '11 at 18:31
  • @kojiro I just read the explanation for 1 is not a prime number. – Vlad Dimitriu Nov 16 '11 at 22:01
  • [2,5] is 3 spaces apart. – Alvin K. Nov 17 '11 at 04:32
  • @Charles: Ah, and every prime is at equal distance between two primes. :-) Thanks. – ShreevatsaR Nov 17 '11 at 08:15
  • @ShreevatsaR that can't be right. I was just trying to say above, 3 is not equidistant from 2 and 5. – kojiro Nov 17 '11 at 21:15
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    @kojiro, 3 is equidistant from 3 and 3. – Gerry Myerson Nov 17 '11 at 22:46
  • @GerryMyerson 3 is only one prime. – kojiro Nov 18 '11 at 02:12
  • @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. – Gerry Myerson Nov 18 '11 at 04:00
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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." – kojiro Nov 18 '11 at 13:09
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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. – Gerry Myerson Nov 19 '11 at 11:02
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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). – Marc van Leeuwen Nov 24 '11 at 12:52
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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.

Marc van Leeuwen
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  • Surely 1 is a prime number? It's only divisible by 1 and itself (1) – m-smith Nov 16 '11 at 14:25
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    1 isn't considered prime because, otherwise, various theorems would be inelagently phrased ("all primes except 1"). – David Mitra Nov 16 '11 at 14:52
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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-number – Ross Millikan Nov 16 '11 at 14:54
  • OK, thanks for clarifying for me :) – m-smith Nov 16 '11 at 17:12
  • 1used to be considered prime; indeed 1 is a necessary factor in "perfect" numbers –  Jul 28 '12 at 12:52
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    @AAA Consider posing questions or providing suitable answers to posted questions in order to gain the 50 reputation points required to begin commenting. (And note that to be perfect 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.) – user642796 Jul 28 '12 at 13:43
  • @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). – Marc van Leeuwen Jul 28 '12 at 15:40
  • @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! – user642796 Jul 28 '12 at 17:06
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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.

Srivatsan
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Alvin K.
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1

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.

*Assuming the Goldbach conjecture

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

Houston
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michaelmross
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