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Let's say that the linear form $ax+by$ represents $n$ if $ax+by=n$ for some positive integer $x$ and $y$.

Call a pair $(a,b)\in\Bbb N\times\Bbb N$ with $\mathsf{gcd}(a,b)=1$ good if, for any $r,s,u,v>1$ with each of $rs,uv,ru,sv,rv,su<(a-1)(b-1)$ (the Frobenius number of $(a,b)$), there is at most one set from among $\{rs,uv\}$, $\{ru,sv\}$ and $\{rv,su\}$ with both components representable by $ax+by$.

Do good pairs exist at all?

If they do, then is it true that for every sufficiently large integer $l$, there is a good pair $(a,b)$ with $a,b\in[l,2l]$?

A bad pair example:

$a=22,b=21,s = 16, t = 17,r = 19,u = 15$

$$10a+4b=rs$$ $$8a+7b=rt$$ $$9a+2b=su$$ $$3a+9b=tu$$

Related Some divisibility constraints in Frobenius coin problem

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    $\begingroup$ A piece of advice. To receive proper attention and get answered, a problem must be presented in a comprehensible way. For your specific question, you can consider, for instance, defining the notion of a "good pair" $(a,b)$, presenting and explaining some examples of good and "bad" pairs, and only then asking whether there is a good pair in any interval $[n,2n]$. $\endgroup$
    – Seva
    Commented Nov 30, 2015 at 20:01
  • $\begingroup$ @Seva Actually I thought about it, I am not sure $n_0$ has a good upper bound at all. $\endgroup$
    – Turbo
    Commented Nov 30, 2015 at 21:04
  • $\begingroup$ @Seva you think it is comprehensible now? I do not have an example for good pairs. $\endgroup$
    – Turbo
    Commented Nov 30, 2015 at 23:08
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    $\begingroup$ Doesn't the Frobenius problem itself immediately imply that the answer to your question is negative? All numbers $n$ greater than $(a-1)(b-1)$ have a representation $n=av+bw$ with $v,w$ positive, so as soon as your $r,s,t,u$ are sufficiently large (or more simply as soon as their products are) you'll inevitably have a positive representation. $\endgroup$
    – Steven Stadnicki
    Commented Dec 1, 2015 at 7:56
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    $\begingroup$ @Turbo You should really clarify the quantifiers in your original question, then - as it currently stands, the natural reading is 'if the condition on the left side of the implication arrow holds, then the condition on the right side holds' (this is what an implication usually means!) which is generally equivalent to saying 'for ALL values which satisfy the condition on the LHS, the condition on the right holds'. $\endgroup$
    – Steven Stadnicki
    Commented Dec 1, 2015 at 8:16

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I don't totally understand your definition of "good" so I don't really know exactly what makes your example bad. I filled some idle time trying to generalize your example of a bad pair. I realize (now) that that is not relevant to your actual question. But having found it anyway, here it is:

Your example of a bad pair $a=2\cdot 11,b=a+1=3\cdot 7$ suggests something like $a=n(4n+3),b=(n+1)(4n-1)$ The rest is nicely patterned as well with (possibly not on the first try) $u,s,t,r=u,u+1,u+2,u+2+n.$ A little fiddling gives

$$a=4n^2+3n,b=4n^2+3n-1,$$ $$s = 4n^2+n-2, t = 4n^2+n-1,r = 4n^2+2n-1,$$$$u = 4n^2+n-3=(n+1)(4n-3)$$

$$(2n^2+n)a+(2n^2-n-2)b=rs$$ $$(2n^2)a+(2n^2-1)b=rt$$ $$(3n+3)a+(4n^2-4n-6)b=su$$ $$(n+1)a+(4n^2-2n-3)b=tu$$

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  • $\begingroup$ Thank you for the time taken. The idea is simple. Lets say a linear form $ax+by$ represents $n$ if $ax+by=n$ for some $x,y\geq0$. Good pair $a,b$ essentially forms a linear form such that if $ax+by$ represents $rs$ and $tu$, it should not represent $rt,su$ and $ru,st$. That is all there is to it. I am looking for an infinite collection of good pair above a certain universal constant $n_0$. $\endgroup$
    – Turbo
    Commented Dec 1, 2015 at 6:54
  • $\begingroup$ In example above we have $a=22,b=21$ represent $rs=19\cdot16$ and $tu=17\cdot 15$ but also representing $rt$ and $su$ so $a,b$ is a bad pair. $\endgroup$
    – Turbo
    Commented Dec 1, 2015 at 7:06

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