Cfrgtkky

For all integers $n>1$ there are positive integers $a,b$ such that $a+b=n$ and such that $a+ab+binmathbb P$.
Tested for all $nleq 1,000,000$. Hopefully, someone can explore and explain the heuristics about this conjecture.

This question is related to:
Any odd number is of form $a+b$ where $a^2+b^2$ is prime
Does every power of two arise as the difference of two primes?
Most even numbers is a sum $a+b+c+d$ where $a^2+b^2+c^2=d^2$
Natural numbers large enough can be written as $ab+ac+bc$ for some $a,b,c>0$
${a+b|a,binmathbb N^+wedge ma^2+nb^2inmathbb P}={k>1|gcd(k,m+n)=1}$
Even numbers has the form $a+b$ where $frac{a^2+b^2}{2}$ is prime
Is every positive integer greater than $2$ the sum of a prime and two squares?

It's about a relation $Rsubseteq mathbb N^m$, a function
$f:mathbb N^mto mathbb N$, and an image of a restriction
$operatorname{Im}(f|R)$.

In Goldbachs conjecture the relation is $p,qinmathbb P$, the function is $(p,q)mapsto p+q$ and the image of the restriction is
$2mathbb Nsetminus{2}$.

Maybe some of the conjectures can be generalized?

This is only a partial answer. Reformulating according to the comment of Crostul, we want to find $AB-1 in mathbb P$. Other than the even prime $2$, all primes are odd, so at least one of $A,B$ must be even. For $A+B=N, Nge 6$, any prime generated will have the form $6mpm 1$. So it is necessary (but not sufficient) that for every $N$, there are some $A,B$ such that $AB-1=6mpm 1$ for the conjecture to be true.

$AB-1=6mpm 1Rightarrow ABequiv (0,2) mod{6}$. Any $Nge 6$ can be split into two addends with that property. The following table lists values of residues $mod{6}$ for $N,A,B$ that satisfy $N=A+Bmod{6}$ and $ABequiv (0,2) mod{6}$ (up to the order of $A,B$).

$$begin{array}{ccc}
N&A&B \
0&0&0 \
&2&4 \ \
1&0&1 \
&3&4 \ \
2&0&2 \
3&0&3 \
&1&2 \
&4&5 \ \
4&0&4 \
5&0&5 \
&2&3 \
end{array}$$

This shows that any $N$ can be split into addends $A,B$ such that $ABequiv (0,2) mod{6}$. In every case, it is possible to obtain addends such that $ABequiv 0 mod{6}$. Interestingly, only for $Nequiv (0,3)mod6 Rightarrow Nequiv 0 mod3$ is it possible to obtain addends such that $ABequiv 2 mod{6}$. This means that $AB-1$ can generate numbers of the form $6m-1$ from any $N$, but can generate numbers of the form $6m+1$ only if $Nequiv 0mod3$. It remains open at this point whether the numbers of the form $6mpm 1$ obtained from a particular $N$ will necessarily feature a prime.

Idea: Let $$a+b+ab =pin mathbb{P}$$

then $$n = {a^2+pover a+1}= a-1+{p+1over a+1}$$

So if we take such $p$ and $a$ that $a+1mid p+1$ and $a<n$ then $b={p-aover a+1}$.

Question here is if such $p$ and $a$ alway exist.