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If we look at the difference between consecutive prime numbers, $p \gt 2$, it always appears to be an even number.

For example, here are the seven consecutive primes starting at the $10^{10th}$ prime.

$p_i = \{252097800623, 252097800629, 252097800637, 252097800667, 252097800737, 252097800743, 252097800839\}$

The differences between the consecutive primes above are $\{6, 8, 30, 70, 6, 96\}$, and are all an even number .

This, of course, is automatic for twin primes since by definition they differ by $2$.

Also, this holds for all balanced primes, A006562 - Balanced primes, since we have $2*p_n = p_{n-1} + p_{n+1}$.

There is a table of such values in A001223 - Differences between consecutive primes on OEIS.

My questions are:

(1) Is it considered a conjecture that the difference between consecutive primes $p \gt 2$ is always an even number?

I wasn't sure if there was some argument regarding Prime Gaps that guarantees such a result and it is easy.

(2) Has this been proven?

Note that I found the Prime Difference Function, but is that the latest?


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Any two primes greater than $2$ are odd. So their difference is...? – 1015 Jan 13 '13 at 18:42
@julien: Duh to me! Can a moderator please delete? Thanks! – Amzoti Jan 13 '13 at 18:44
No need to's a cute question! ;-) – amWhy May 21 '13 at 0:13
up vote 14 down vote accepted

You are thinking way too hard about this.

First, to the question in the title (but not as asked in the text) no: $3-2=1$.

As asked in the text for odd primes, then the difference between two odd numbers is always an even number: $$2p+1 - (2q+1) = 2(p-q).$$

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In other words, the specific case of consecutive odd primes having an even difference follows from the generality (that two odd numbers have an even difference). – hardmath Jan 13 '13 at 18:46

Think binary. For all odd primes the last digit, in binary, is 1. So, for any two odd primes:

P1 = ...1 P2 = ...1 X = P1 - P2 = (+/-)...0

Where ... represents whatever leading digits there are.

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I upvote when I see a total negative amount of votes and the answer is correct. +1 – mick Sep 1 '13 at 16:46

Yes, always. If we get two prime numbers $\gt 2$, then both are odd and difference two odd numbers is always even. This can be simply proven:
Each prime number $\gt 2$ is odd. This can be proven by contradiction: If those number is even, then this number is divisible by $2$ and then it is not a prime number.
Difference two odd numbers is even always. Let us have some two odd numbers $p_1$ and $p_2$. These numbers are odd and according to this we can express these numbers as $p_1=2k+1$ and $p_2=2l+1$, where $k\ge 1$ and $l \ge k$, then $\Delta = p_2 - p_1 = 2l+1-2k-1=2(l-k)$. This number is even and difference two odd, including prime, numbers($\gt 2$) is always even number.

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