# Is the difference between consecutive prime numbers always an even number?

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?

Regards

• Any two primes greater than $2$ are odd. So their difference is...? Commented Jan 13, 2013 at 18:42
• @julien: Duh to me! Can a moderator please delete? Thanks! Commented Jan 13, 2013 at 18:44
• No need to delete...it's a cute question! ;-) Commented May 21, 2013 at 0:13

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: $$2p+1 - (2q+1) = 2(p-q).$$

• 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). Commented Jan 13, 2013 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.

• I upvote when I see a total negative amount of votes and the answer is correct. +1
– mick
Commented Sep 1, 2013 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.

Yes this is absolutely true ,Difference between two consecutive or any prime numbers will always an even number. iff number is not 2 & 3.

Proof argument logic, There is common fact that the difference between any two odd numbers must be an even number & any prime number must be odd number so difference between any prime number other than 2,3 will be even number. Thanks Anil Kesarwani