All other prime gaps (except between $2$ and $3$) are even. How can one show that except for the first prime gap (between $2$ and $3$), all prime gaps are even? 
Prime gap means the difference of two consecutive primes.
For example: $5-3=2$ (even) or $17-13=4$ (even) or $23-19=4$ (even).
I have no idea how to start.
 A: Since all prime numbers except $2$ are odd, then if a prime gap $g$ between primes $p_1,p_2$ would be odd, then $p_2=p_1+g$ would be the sum of two odd numbers which is even, and since the only prime which is even equals $2$, thus we get the desired contradiction.
A: All primes other than $2$ are odd, hence gap between them is even (as difference of two odd numbers is always even).
A: $1$. The only even prime is $2$ and all other are odd primes.
$2$. Difference between two odd numbers is always even.
First posted as hint but now @Cato asked for a proof so here We go:
Proof of $1$:
Suppose that there is another prime which is even (other than $2$), then it must be of the form $2k$ where $k$ being a positive integer, so that prime number can be written as $2 \times k$ which is enough to say that it is composite (Or not prime). A contradiction.
Proof of $2$. 
As known (I don't think it also need a proof), every odd number is of the form $2k+1$, where $k$ is an integer. We just have to calculate the difference b/w two odd primes (Remember the word odd) so let first one be $2k_1+1$and second one be $2k_2+1$. Difference $= (2k_2+1-(2k_1+1))= 2k_2-2k_1=2(k_2-k_1)=$ even number.
I think it is enough. :) :) 
