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Is every sufficiently large positive integer $A$ of the form $ab + ac + bc + 1$

where $a,b,c$ are some positive integers larger than some given positive integer $d$ ?

How large is sufficiently large , in other words how do we know our number $A$ is sufficiently large ( as a function of $d$ ) ?

Update :

I read the following about positive integers that are not of the form $ ab + ac + bc $:

https://oeis.org/A000926

And I quote about that sequence above : " Note that for n in this sequence, n+1 is either a prime, twice a prime, the square of a prime, 8 or 16 "

This is why I considered the weird +1 in $ab + ac + bc + 1$

So composites larger than 16 not a prime , twice a prime or the square of a prime are of the form $ab + ac + bc + 1$.

And I wonder why !

I guess that relates strongly to the Original posted problem. Maybe I should have mentioned that before , sorry.

Will Jagy's answer gave the same OEIS link which I already knew. This also explains some of the comments.

Im no expert in genus theory or Galois theory but I assume this has a simple answer.

mick
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  • I don't think you can determine "sufficently large" as a function of $d$, since $a,b,c$ can be arbitrarily larger than $d$. Whatever function $f(d)$ you determine, there's nothing stopping $a=f(d)+1$. – KSmarts Dec 19 '14 at 22:48
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    What is the reason for the $+1$? It seems simpler to ask about numbers of the form $ab+ac+bc$, and is clearly equivalent... – Micah Dec 19 '14 at 22:56
  • @Micah I was thinking about writing composites in the form ab + ac + bc + 1 ... but you are right I guess. – mick Dec 19 '14 at 23:00
  • I updated the question. Really wonder why almost every composite is of the form ab + ac + bc + 1 See the update. Will probably accept as answer once that is explained. – mick Dec 22 '14 at 23:30

2 Answers2

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Euler's "numerus idoneus" (or "numeri idonei", or idoneal, or suitable, or convenient numbers).

https://oeis.org/A000926

It is conjectured that the list given here is complete. Chowla showed that the list is finite and Weinberger showed that there is at most one further term.

(6) n is not of the form ab+ac+bc with 0 < a < b < c. [Rains]

Will Jagy
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Then equation: $$XY+XZ+YZ=N$$

If we ask what ever number: $p$

That the following sum can always be factored: $p^2+N=ks$

Solutions can be written. $$X=p$$ $$Y=s-p$$ $$Z=k-p$$

I like this formula the solutions of this equation. Number of solution for $xy +yz + zx = N$

Community
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individ
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  • +1 I looked at your profile, you certainly like small diophantine equations ! I was fascinated. I edited my question, maybe you can adress the update with a new answer ? – mick Dec 20 '14 at 22:09