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.