Small geometric progression modulo N An problem related to integer factorization using the General Number Field Sieve is the following:  

Let $N$ be a composite. Must there exist a 5-term geometric progression $\lbrace a_0,a_1,a_2,a_3,a_4\rbrace$ (mod $N$) such that each term is $O(N^{2/3})$?
  We also require that  determinant
  $\bigl(\begin{smallmatrix}
a_4&a_3&a_2\\\\
a_3&a_2&a_1\\\\
a_2&a_1&a_0
\end{smallmatrix} \bigr)$ is not zero.
  What is known about such geometric progressions?

Some background (not needed to answer the question): For factorizations using the Number Field Sieve, one would construct 2 polynomials $f(x)$ and $g(x)$ such that $f(m)\equiv g(m) \equiv 0$ (mod $N$) for some $m$. Usually, one of the polynomial will be of degree 1, the other ranging from 2 to 8 for practical purposes. There exists methods to construct 2 polynomials of similar degree, such that both are non-linear. The problem stated essentially gives one a way to construct 2 cubic polynomials. This method may have some advantages over the traditional linear case. 
Note that the problem do not ask how can one find the actual geometric progression, as that is believed to be hard and at the same time an open problem.  
Motivation:
I am posting this problem as the sources I read suggests that such series do exist by "counting argument". The part I am wondering about is: how does the counting argument work? 
Initial Thoughts
If I may humbly present some of my naive ideas:
We start off with $2*N^{2/3}$ possibilities for the first term $a$.
Next, since we seek a second term $b$ bounded by $O(N^{2/3})$, we similarly allow $2*N^{2/3}$ possibilities for it. However, this fixes the common ratio $r=ba^{-1}$ (mod $N$), whence we have no control over the 3rd, 4th and 5th terms.
We reason that such an $r$ must exist for each pair of $\lbrace a,ar \rbrace$, as failure would imply that $a^{-1}$ does not exist, which implies that we have found a factor of $N$.  
If we were to assume that the effect of multiplying by $r$ (mod $N$) results in a random permutation of the integer, the 3rd, 4th and 5th terms each have $(2*N^{2/3})/(2*N)=N^{1/3}$ chance of being bounded by $O(N^{2/3})$.
This gives us a grand total of $2*N^{2/3}*2*N^{2/3}/(N^{1/3})^3=4*N^{1/3}$ valid series.  
Suppose this line of thought holds, the difficult part is to account for determinant not zero. (This is stated as to ensure the series is not a second-order linear recurrence over $\mathbb{Q}$, which is a requirement for the Number Field Sieve to work.) It is clear that if the series is of the form $\bigl(\begin{smallmatrix}
ar^4&ar^3&ar^2\\\\
ar^3&ar^2&ar\\\\
ar^2&ar&a
\end{smallmatrix} \bigr)$
with $ar^4 < N^{2/3}$, such that no modulo reduction happens, the determinant is zero by default. So in some sense the series "must exceed $N$ at some point". My guess is that if we remove all series without the modulo reduction then the remaining probability presumably will tell us that the geometric progression should exist.  
Is my argument sound? How does one go about counting the series to remove?
 A: Here is the abstract for Aishwarya A. Vardhana, Small geometric progressions modulo n for deterministic polynomial selection, a prize-winning entry in the 2012 Intel Science Fair (Vardhana was in high school at the time): 
In this project we present a solution to an unsolved problem known as "Small Geometric Progressions Modulo $N$" proposed by Dr. Peter L. Montgomery in 1993. Montgomery’s problem postulates the creation of two irreducible cubic polynomials with properties ideal for the success (i.e. factorization) of the General Number Field Sieve (GNFS); asymptotically the fastest algorithm for factoring a large integer $N$ with no small prime factors, such as an RSA modulus. The first step of the GNFS, known polynomial selection, sieves through random polynomials to find one or two polynomials with decent yield. We propose a deterministic method for creating two polynomials with ideal properties such as small rational coefficients, a common root $m$, and no common factor. These polys will produce the most smooth numbers (i.e. numbers with small prime factors) in the shortest amount of time thereby reducing the overall computation time of the GNFS. Using higher level modular arithmetic, Linear Algebra, and Number theory we were successful in solving the Montgomery's problem. The solution is a deterministic algorithm in polynomial time, much more efficient and flexible in comparison to the traditionally used Kleinjung's method.
I found this at http://www.neohstem.org/assets/pdf/Society%20for%20Science%20&%20the%20Public%20-%20Math.pdf but I haven't found the actual project anywhere, and I don't know whether she actually solves the problem we're talking about here. I don't know whether she ever published anything on it. 
