## Hard divisibility (sort of diophantic) problem

Short intro: $\lambda_1$, $\lambda_2$ and $\lambda_3$ are (different!) eigenvalues of a $S$ matrix from knot theory and (for my $S$ matrices and what reason ever) always have the form $\pm q^i$ with $i \in Z$. (Or $\pm q^i*r^j*\cdots$ but one free variable is bad enough...) o is a quantum dimension resp. unknot value and always is a Laurent polynomial (a sum $a_j*q^j$, $j \in Z$), also for what reason ever. The value of o below follows from assuming a cubic skein relation.
$$o=\frac{-1+\lambda_1^3\lambda_2^3\lambda_3^2+\lambda_1^3\lambda_2^2\lambda_3^3+\lambda_1^2\lambda_2^3\lambda_3^3-\lambda_1^6\lambda_2^5\lambda_3^5-\lambda_1^5\lambda_2^6\lambda_3^5-\lambda_1^5\lambda_2^5\lambda_3^6+\lambda_1^8\lambda_2^8\lambda_3^8}{\lambda_1^3\lambda_2^3\lambda_3^3(-\lambda_1-\lambda_2-\lambda_3+\lambda_1^2\lambda_2^2\lambda_3+\lambda_1^2\lambda_2\lambda_3^2+\lambda_1\lambda_2^2\lambda_3^2)}$$

A bit of experimenting will show only few $\lambda_1=\pm q^i$, $\lambda_2=\pm q^j$, $\lambda_3=\pm q^k$ will give a Laurent polynomial (in q) for o (the rest has accumulating trash in the denominator). (A working sample input: $\lambda_1=q^3$, $\lambda_2=-q^11$, $\lambda_3=-q^{-9}$.)
Obviously I would dance with joy if only a finite number of valid $\{i,j,k\}$ sets exist (which does not hold, but maybe most sets follow a formula and only finite many exception exist). Maybe it reduces the list when I add that some o1 must be a Laurent too?
o1=((-1+\lambda_2*\lambda_3)*(1+\lambda_2*\lambda_3)*(-1+\lambda_1*\lambda_2*\lambda_3)*(1+\lambda_1*\lambda_2*\lambda_3)*(1+\lambda_1^2*\lambda_2^2*\lambda_3^2)*(-1+\lambda_1^3*\lambda_2^3*\lambda_3^2)*(1-\lambda_1*\lambda_2^2*\lambda_3+\lambda_1^2*\lambda_2^4*\lambda_3^2)*(-1+\lambda_1^3*\lambda_2^2*\lambda_3^3)*(1-\lambda_1*\lambda_2*\lambda_3^2+\lambda_1^2*\lambda_2^2*\lambda_3^4))/(\lambda_1^4*\lambda_2^6*\lambda_3^6*(-\lambda_1+\lambda_2)*(-\lambda_1+\lambda_3)*(-\lambda_1-\lambda_2-\lambda_3+\lambda_1^2*\lambda_2^2*\lambda_3+\lambda_1^2*\lambda_2*\lambda_3^2+\lambda_1*\lambda_2^2*\lambda_3^2)^2)
And o2 and o3 (cyclic change of \lambda_1,\lambda_2,\lambda_3) too, of course.

Even if the formulae were less nightmarishly long, I haven't the slightest idea whether one can attack this problem (other than with brute force: make a list of working {i,j,k} and search for patterns). (OK, one slight idea: write everything via quantum integers and seek for quantum binomial coefficients?)
Do you have some input on finding all allowed sets {i,j,k}? (If not, nevermind, then I go back to good old diagram drawing :-)

EDIT: Quantum -> try q=1 limit -> signature +++ is useless since the limit is o=0 then, signature +-- leads to $i|j+k$ and $j-k|(i+j+k)*(2*i+j+k)*(2*i+3*j+3*k)$. (Other signatures of $\{\lambda_1,\lambda_2,\lambda_3\}$ are equivalent.)

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Unreadable. Ever hear of TeX? – Gerry Myerson Jul 6 2011 at 12:41
Yes, but...even more unreadable WITH TeX - I tried, the formula goes over the whole page and I don't know TeX linebreaks. Moreover, only the first formula converted - as I put  around the second, it didn't. Maybe I should UL a screenshot of the MATHEMATICA output instead? I'm a total TeX n00b. – Hauke Reddmann Jul 6 2011 at 13:46
Just to make sure that I get it right - we're supposed to decipher this mess, guess the correct definitions and think about the problem when you're not even willing to work on a decent way of presenting your question? You should indeed dance with joy when you get a response. – Franz Lemmermeyer Jul 6 2011 at 15:43
I see the question as similar to the following one: Given the equation a= b+c and the expression q^z + q^w, are there only finitely many z and w, and how do I find them? For me, the sensible answer is: In the question, I see only two z's and two w's, so of course there are only finitely many of them. You can use grep to find them. Otherwise, I have no idea what he is asking about, even after setting aside issues like understanding the terminology. Which is a shame, because he used to ask understandable questions. Gerhard "Questioning the Questions of Questioners" Paseman, 2011.07.06 – Gerhard Paseman Jul 6 2011 at 18:17
I tried to understand by fixing the TeX, but I surrender. Hauke, if you can make it more readable (and be clear about what exactly you are asking), you have an audience. – Kevin O'Bryant Jul 7 2011 at 0:49
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