Dealing with Primes

We know that $a-b$ and $a+b$ are either both odd or both even.

They are primes $\Rightarrow$ They both are odd. $\therefore$ One of $a,b$ is odd and the other is even.

$2$ is the only even prime.

$\therefore\space b\space =\space 2$ .

Hence, $a-2,\space a,\space a+2$ are consecutive odd primes. Hence they are $3,\space 5,\space 7$ .

$5,\space 2,\space 5-2,\space 5+2$ are all prime.

Sum is $2+3+5+7=17$ , which is a prime.

All the primes are odd numbers except one which is 2.
In the instant case, since the sum as well as the difference of two primes are also prime numbers, it is necessary that one of the prime numbers is even i.e. 2 otherwise, their sum and difference would result into two even numbers and hence not prime. Let the other prime, being odd number, be (2n+1).
The remaining two primes would be,
(2n+1) - 2 = (2n - 1) and,
(2n+1) + 2 = (2n+3)
Arranging them in ascending order, the 4 primes are :
1. 2,
2. (2n - 1) = {2(n - 2) + 3},
3. (2n + 1)= {2(n - 1) + 3},
4. (2n + 3) = {2(n) + 3},
Since (n - 2), (n - 1) and n are three consecutive numbers, one of the prime has to be a multiple of 3. Since the only prime which is a multiple of three, is 3 itself, the smallest odd prime is 3, hence we have,
2n - 1 = 3 or 2n = 4, so n = 2, The remaining two prime numbers are
2n + 1 = 5, and
2n + 3 = 7,
So, sum of 4 prime numbers
= 2+3+5+7 = 17 .