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I am trying to solve the equation system eq1==eq2 for $\sigma1$ and $\sigma2$ where eq1 and eq2 are functions of 5 positive real variables as defined below:

eq1=(2 Sqrt[5 + 4 x1 + x1^2 + 2 y1 + y1^2])/(Sqrt[5] (2 + x1) (1 +y1))

eq2= (2 (1 + s) Sqrt[5 + 5 s^2 + 4 x1 \[Sigma]1 + x1^2 \[Sigma]1^2 + 2 y1 \[Sigma]2 + y1^2 \[Sigma]2^2 + 2 s (5 + 2 x1 \[Sigma]1 + y1 \[Sigma]2)])/(Sqrt[5] (2 + 2 s + x1 \[Sigma]1) (1 + s + y1 \[Sigma]2))

I am interested only in those solutions which are only a function of variable s. As a matter of fact I can simply verify that s+1 is a solution:

eq2 /. {\[Sigma]1 -> 1 + s, \[Sigma]2 -> 1 + s} //FullSimplify[#, s > 0] &

however, I have not been able to find this solution using Mathematica.

I have tried the following

Reduce[eq1 == eq2 && 0 < x1 && 0 < y1 && 0 < \[Sigma]1 &&0 < \[Sigma]2 && 0 < s, {\[Sigma]1, \[Sigma]2},Reals] // FullSimplify

and

Solve[eq1 == eq2, {\[Sigma]1, \[Sigma]2}, Reals] // FullSimplify

but none of these result in the simple solution s+1. My question is what is the right approach to solving this kind of equations in Mathematica.

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1 Answer 1

SolveAlways might be the way to go. You are after relations between $\sigma1$, $\sigma2$ and $s$ which make the equation hold for all values of $x1$ and $y1$, so you would use:

SolveAlways[eq2 == eq1, {x1, y1}]

Unfortunately this seems very slow and I wasn't patient enough to let it finish. However, squaring both sides gives results quickly:

SolveAlways[eq2^2 == eq1^2, {x1, y1}] // DeleteDuplicates

(* {{σ1 -> 0, s -> -1}, {σ1 -> 1 + s, σ2 -> 1 + s}, {σ2 -> 0, s -> -1}} *)
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Many thanks @Simon Woods. I am still running the first line to see if I get a solution. –  Mikael Anderson Jun 5 at 10:01

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