# Do there exist infinitely many positive integers which cannot be represented by $ab+bc+ca$?

A friend of mine taught me the following question which he's been trying to solve. We are facing difficulty.

Question : Two sets $A,B$ are defined as $$A=\{n\ |\ n\ \text{is a positive integer}\},$$ $$B=\{ab+bc+ca\ |\ a,b,c\ \text{are positive integers}\}.$$ Then, is $|A\cap B^c|$ infinite where $B^c$ is the complement of $B$?

What we've found is the followings : Let $m$ be a positive integer.

• $2m+1\in B$. Take $(a,b,c)=(1,1,m).$

• $4m+4\in B$. Take $(a,b,c)=(2,2,m)$.

• $12m+2\in B$. Take $(a,b,c)=(1,2,4m)$.

• $60m-14\in B$. Take $(a,b,c)=(3,2,12m-4)$.

• $60m+6\in B$. Take $(a,b,c)=(3,2,12m)$.

• $60m-6\in B$. Take $(a,b,c)=(4,6,6m-3)$.

• $60m-26\in B$. Take $(a,b,c)=(4,6,6m-5)$.

Also, we've found that $$1,2,4,6,10,18,22,30,42\in{A\cap B^c}.$$

However, we don't know what to do next. Can anyone help?

• You might want to use something other that $B'$ to mean the complement, as the $'$ is easy to overlook (and thus changes the meaning). An alternative is to write $\mathbb{N}\setminus B$. – Tobias Kildetoft Nov 13 '14 at 11:09
• Or even simpler $B^c$ – Alexandre Halm Nov 13 '14 at 11:18
• @TobiasKildetoft and Alex H.: Thank you. I edited it. – mathlove Nov 13 '14 at 11:22
• The formula there. math.stackexchange.com/questions/419766/… Having the formula, you can further decide which numbers to choose. And find out if there are any other numbers. – individ Nov 13 '14 at 11:49

Borwein and Choi showed in a short paper in 2000 that there are 18 or 19 solutions, and the first 18 are

$$1, 2, 4, 6, 10, 18, 22, 30, 42, 58, 70, 78, 102, 130, 190, 210, 330, 462.$$

(This is sequence A025052 in OEIS.) As of that paper, it was known that the other solution, if it exists, is $> 10^{11}$, and that no such solution exists if the Generalized Riemann Hypothesis is true. I do not know whether this outstanding case has since been resolved some other way. Borwein and Choi's proof begins by splitting into the squarefree and nonsquarefree cases (the latter case gives only $4$ and $18$), and then whittles down the possibilities in part by producing some special cases not unlike your list above.

Interestingly, this problem turns out to be equivalent to the classification of discriminants for which no indecomposable positive definite binary quadratic forms exist.

• Thank you very much for the information! – mathlove Nov 14 '14 at 9:18
• You're welcome, I hope you found it useful! – Travis Nov 14 '14 at 10:28

Admittedly, I can't prove or disprove this, but if I had to prove/disprove this, I'd make the assumption that this set would be finite.

Then there is some value $n_0$ for which all values in $B^C < n_0$.

Then you'd either have to construct a new number $n_1$ in $B^C$ larger than $n_0$ that doesn't hold, or show a way to construct $a, b, c$ for such number.

But probably that's not simple.