# For all $n>1$ there are positive $a+b=n$ such that $a+ab+b\in\mathbb P$

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.

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?

• 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 5 '18 at 20:06
• 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 5 '18 at 20:45
• 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 5 '18 at 21:44
• Where did this conjecture come from? What is special about the expression $a+ab+b$? – Servaes Dec 5 '18 at 21:49

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.

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 $$a then $$b={p-a\over a+1}$$.

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

• Nice idea, but far from an answer. Or am I missing something obvious? – Servaes Dec 5 '18 at 20:12
• -1 This is a comment, not an answer. – Servaes Dec 5 '18 at 20:50