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For all integers $n>1$ there are positive integers $a,b$ such that $a+b=n$ and such that $a+ab+b\in\mathbb P$. Tested for all $n\leq 1,000,000$. 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 $a,b,c>0$
$\{a+b|a,b\in\mathbb N^+\wedge ma^2+nb^2\in\mathbb P\}=\{k>1|\gcd(k,m+n)=1\}$
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 $R\subseteq \mathbb N^m$, a function $f:\mathbb N^m\to \mathbb N$, and an image of a restriction $\operatorname{Im}(f|R)$.
In Goldbachs conjecture the relation is $p,q\in\mathbb P$, the function is $(p,q)\mapsto p+q$ and the image of the restriction is $2\mathbb N\setminus\{2\}$.

Maybe some of the conjectures can be generalized?

Lehs
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    Equivalently, you conjecture that for every positive integer $n>1$ there exists a positive integer $a<n$ such that $$a+a(n-a)+(n-a)=-a^2+na+n,$$ is prime. – Servaes Dec 05 '18 at 20:06
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    If we put it in other terms you are claimin for all $n$ there is a prime equal go $n + (n-a)a$ for some $a$. – fleablood Dec 05 '18 at 20:45
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    Calling $A=a+1$, and $B=b+1$, this is equivalent on saying that for all $N$ there exist positive integers $A,B \ge 2$ such that $A+B=N$, and $AB-1$ is prime. – Crostul Dec 05 '18 at 21:44
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    Where did this conjecture come from? What is special about the expression $a+ab+b$? – Servaes Dec 05 '18 at 21:49

2 Answers2

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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 $A+B=N,\ N\ge 6$, any prime generated will have the form $6m\pm 1$. So it is necessary (but not sufficient) that for every $N$, there are some $A,B$ such that $AB-1=6m\pm 1$ for the conjecture to be true.

$AB-1=6m\pm 1\Rightarrow AB\equiv (0,2) \mod{6}$. Any $N\ge 6$ can be split into two addends with that property. The following table lists values of residues $\mod{6}$ for $N,A,B$ that satisfy $N=A+B\mod{6}$ and $AB\equiv (0,2) \mod{6}$ (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 $AB\equiv (0,2) \mod{6}$. In every case, it is possible to obtain addends such that $AB\equiv 0 \mod{6}$. Interestingly, only for $N\equiv (0,3)\mod6 \Rightarrow N\equiv 0 \mod3$ is it possible to obtain addends such that $AB\equiv 2 \mod{6}$. 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 $N\equiv 0\mod3$. It remains open at this point whether the numbers of the form $6m\pm 1$ obtained from a particular $N$ will necessarily feature a prime.

Keith Backman
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Idea: Let $$a+b+ab =p\in \mathbb{P}$$

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

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

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

nonuser
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