TITLE: Any odd number is of form $a+b$ where $a^2+b^2$ is prime
QUESTION [50 upvotes]: This conjecture is tested for all odd natural numbers less than $10^8$:  
If $n>1$ is an odd natural number, then there are natural numbers $a,b$  such that $n=a+b$ and $a^2+b^2\in\mathbb P$.  
$\mathbb P$ is the set of prime numbers.
I wish help with counterexamples, heuristics or a proof.

Addendum: For odd $n$, $159<n<50,000$, there are $a,b\in\mathbb Z^+$ such that $n=a+b$ and both
$a^2+b^2$ and $a^2+(b+2)^2$ are primes.

As hinted by pisco125 in a comment, there is a weaker version of the conjecture:

Every odd number can be written $x+y$ where $x+iy$ is a Gaussian
  prime.

Which give arise to a function:
$g:\mathbb P_G\to\mathbb O'$, given by $g(x+iy)=x+y$, where $\mathbb O'$ is the odd integers with $0,\pm 2$ included. 
The weaker conjecture is then equivalent with that $g$ is onto.  
The reason why the conjecture is weaker is that any prime of the form $p=4n-1$ is a Gaussian prime. The reason why $0,\pm 2$ must be added is that $\pm 1 \pm i$ is a Gaussian prime.

REPLY [4 votes]: One equivalent way to formulate this conjecture is also the following:
For each natural number $n$ we have:
$ \displaystyle 0 = \prod_{\underset{\gcd(a,b)=1}{2n+1=a+b}} (\Omega(a^2+b^2)-1)$
This is equivalent as to say that the polynomial $\displaystyle f_n(t) = \prod_{\underset{\gcd(a,b)=1}{2n+1=a+b}} (t-\Omega(a^2+b^2))$ has zero $1$.
where $\Omega$ counts the prime divisors with multiplicity.