This is a followup to this question, thanks to Yong Hao Ng for his help with that. I understand the proof now, cleaned it up and tweaked it a little bit. Very helpful.
That problem concerned the three-variable case. For various reasons, the answer to the four-variable case is the same. Now comes the five variable case. If the technique from the previous problem generalizes, I cannot yet see how.
Consider the following nonlinear modular system of equations in five variables and a parameter X:
(a2-X)[(a3+1)(a4+1)(a5+1)-1] = 0 mod a6
(a3-X)[(a4+1)(a5+1)(a6+1)-1] = 0 mod a2
(a4-X)[(a2+1)(a5+1)(a6+1)-1] = 0 mod a3
(a5-X)[(a2+1)(a3+1)(a6+1)-1] = 0 mod a4
(a6-X)[(a2+1)(a3+1)(a4+1)-1] = 0 mod a5
I conjecture this system has no solution for X=6, 6 < a2 < a3 ... < a6 for ai even. Consistent with observations, can't find a counterexample. I have a solution for X=10, conjecture for even X >= 10 there are infinitely many solutions. Solutions with X=8 aren't of interest. Solutions with X=6 would be.
Proof or counterexample would be the best outcome, but a bound on any one of the ai with X=6 would be almost as good, because then the question could be answered computationally.
I've tried the technique of 1 below (let d = gcd of the two highest numbers, for the remaining ai express Xai as a multiple of d, get a new system based on a mutually prime pair of moduli) but with more than three variables I cannot see how to show that for the remaining ai, Xai is a multiple of d (although it very well may be true).
Tried a few different approaches, but every time I get stuck in the complexity of having so many variables. (ai+1)(aj+1)(ak+1)-1 is the sum of all products of length 1 to 3 that can be formed from aj, aj and ak, which makes things complicated. Grateful for any suggestions, strategies to try, etc.
--BF