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I am currently working on a game-theoretic paper and trying to match two complex best response functions. The equation system includes two decision variables x and y, for which I want to find all possible solutions, and two static parameters b and t. The best response functions are given by

x == 1/18 (-6 b - t + 12 y + Sqrt[t] Sqrt[12 b + t + 12 y]

and

y == 1/18 (-6 b + t + 12 x - Sqrt[t] Sqrt[-12 b + t - 12 x])

Using

Solve[x == 1/18 (-6 b - t + 12 y + Sqrt[t] Sqrt[12 b + t + 12 y]) && 
      y == 1/18 (-6 b + t + 12 x - Sqrt[t] Sqrt[-12 b + t - 12 x]), {x, y}]

yields:

{{x -> -b, y -> -b}, {x -> 1/25 (-25 b - 2 t), y -> 1/225 (-225 b - 17 t)}}

I also tried the Reduce command with the same result. Working on, two problems occurred.

Problem 1: Checking results

I wanted to double check my results and therefore replaced x and y in both equations above with

x = 1/25 (-25 b - 2 t) 
y = 1/225 (-225 b - 17 t)

and used the Simplify command, to see whether the results hold true. While this was the case for the first equation, simplifying the second equation yielded

t==0.

Thus, the solution seems valid only for t = 0, though I expected Solve to give me generic solutions for any value of t (in fact, in my model t > 0).

Problem 2: Further solutions

I also inverted the second equation using

Solve[y == 1/18 (-6 b + t + 12 x - Sqrt[t] Sqrt[-12 b + t - 12 x]), x]

which yielded

x == 1/8 (4 b - t + 12 y - Sqrt[t] Sqrt[-8 b + t - 8 y])

and

x == 1/8 (4 b - t + 12 y + Sqrt[t] Sqrt[-8 b + t - 8 y])

Solving the equation system as before but using the inverted form of the second equation yielded different results

{{x -> -b, y -> -b}, {x -> 1/225 (-225 b + 17 t), y -> 1/25 (-25 b + 2 t)}

Again, I tried to check these results by inserting them into both equations, with the same result as before: True for the first equation but t==0 for the second one.

Therefore my overall question is:

How can I get all possible and reliable solutions for the equation system above?

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  • $\begingroup$ Seems to be due to a weak spot in the solution verifier when parameters and radicals are both present. Will have a closer look. $\endgroup$ – Daniel Lichtblau Nov 14 '14 at 19:09
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Probably I do not quite understand the problem, but if you want solutions which would be valid for all t, then I would first eliminate it from the equations above. Unfortunately SolveAlways[] don't work in this case. Still You can calculate GroebnerBasis and eliminate t like this

gb = GroebnerBasis[{x == 
    1/18 (-6 b - t + 12 y + Sqrt[t] Sqrt[12 b + t + 12 y]), 
   y == 1/18 (-6 b + t + 12 x - Sqrt[t] Sqrt[-12 b + t - 12 x])}, {x, 
   y}, {t}]

And then solve obtained equations

Solve[(# == 0) & /@ gb, {x, y}]

{{y -> -2 b - x}, {x -> -b, y -> -b}}

This gives relation between y and x, which should be valid for any t.

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