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While I was working on a paper on graph theory, I encountered a problem which I think is a number-theory-problem. I don't know if there are any tools to answer the question.

Find all natural numbers $n$, or prove there are infinitely many $n$, such that the equation $ab+bc+ca=n$ has no answer in $\mathbb{N}$.

Can help me or introduce some tools to answer this question?

Thanks

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    $\begingroup$ Setting $a = 0$ we always get a solution (if this is allowed). If not, setting $a = 1$ it is not that hard to see how to get a solution whenever $n+1$ is not prime. I don't know if there is a way to find solutions for $n+1$ prime yet... (edit: For clarity: If $a = 1$, we have $n+1 = (b+1)(c+1)$, this has always a solution if $n+1$ is not prime). $\endgroup$
    – Dirk
    Sep 27, 2017 at 11:39
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    $\begingroup$ oeis.org/A025052 (the remark "probably the list is complete" means that your question "probably" will not be settled here). $\endgroup$ Sep 27, 2017 at 11:45
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    $\begingroup$ Related discussions at mathoverflow.net/questions/6388/… and mathoverflow.net/questions/33411/… $\endgroup$ Sep 27, 2017 at 13:28
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    $\begingroup$ Actually, the OEIS entry comments that According to Borwein and Choi, if the Generalized Riemann Hypothesis is true, then this sequence has no larger terms, otherwise there may be one term greater than $10^{11}$. So, while there is uncertainty about the full list, there are definitely only finitely many. $\endgroup$ Sep 27, 2017 at 13:33

1 Answer 1

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This is an elaboration of Emil Jeřábek's important comment, and contains no original contribution. The OP's problem was examined in depth by Borwein-Choi (1999), and their article is available for free here. I will summarize the content of this article below.

Let us consider an integer $n\geq 2$ that cannot be written as $ab+bc+ca$ with integers $a,b,c\geq 1$. Using Lemma 1.1 with $k=1$, we see that $n$ is even. Using Theorem 2.6, it follows that either $n\in\{4,18\}$ or $n=2p_1\dots p_r$ with distinct odd primes $p_j$. The proof of these two results are elementary but highly nontrivial. Then, using some results of Andrews and Crandall, the authors deduce that the class number $h(-4n)$ equals $2^r$ (which is the number of genera of discriminant $-4n$). It is classical that $h(-4n)$ is of size $n^{1/2+o(1)}$ while $2^r=n^{o(1)}$, hence the list of exceptional $n$'s is certainly finite. In fact, Weinberger (1973) analyzed the condition $h(-4n)=2^r$ carefully, and this way we know that either $$n\in\{2,4,6,10,18,22,30,42,58,70,78,102,130,190,210,330,462\},$$ or $n$ is possibly a further single number beyond $10^{11}$. The last possibility can only occur if the Generalized Riemann Hypothesis (GRH) fails for the $L$-function of some quadratic Dirichlet character.

To summarize, there are $18$ exceptional integers $n\geq 1$ if GRH holds, and possibly one further exceptional $n\geq 1$ if GRH fails. (In the above paragraph I restricted to $n\geq 2$ for convenience.)

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