Every number $2^N > 4$ can be expressed by the sum of 2 primes? For example the first cases are:
$2^3= 8 = 3+5$
$2^4= 16 = 3+13$
and so on ...

32=2^5.It is of the form 4k.we know that primes of the form 4a+1,4b-1. Now we write  32=4a+1+4b-1.then a+b=8. By solving a+b=8, we get the pairs (a,b)=(3,5),(7,1) for which 4a+1,4b-1 are primes. In this way we can find 2 primes for every power  2. For ex. 22 is of the form 4k+2. Now we write 22= 4a+1+4b+1where 4a+1 and 4b+1 are primes now we get a+b=5 For (a,b)=(1,4),we get 22=5+17.now we write 22=4a-1+4b-1 From this we get a+b=6For (a,b)=(1,5),(3,3) we can write 22=3+19=11+11.If N =4n=p+q where p, q are primes of the form 4a+1 and 4b-1 If N=4n+2=4a+1+4b+1=4p-1+4q-1 where 4a+1,4b+1,4p-1,4q-1 are primes.  For given any even integer N, we can  find 2 primes as the sum is N.This is an elementary approach to write  for any given even integer as a sum of 2 odd primes.
