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?

  • 1
    $\begingroup$ 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. $\endgroup$ – Servaes Dec 5 '18 at 20:06
  • 2
    $\begingroup$ 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 $a$. $\endgroup$ – fleablood Dec 5 '18 at 20:45
  • 3
    $\begingroup$ 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. $\endgroup$ – Crostul Dec 5 '18 at 21:44
  • 2
    $\begingroup$ Where did this conjecture come from? What is special about the expression $a+ab+b$? $\endgroup$ – 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 $a<n$ then $b={p-a\over a+1}$.

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.