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I have a rather large rational expression exp involving real variables $x_1, x_2, \dots, x_{10}$. The expression is of the following form

 exp = (64*x1^4*x7^4 + 64*x2^4*x7^4 + 16*x2^2*x7^2*(4*x3*x7 + x5 - 2*I*x9)*(x6 - 2*I*x9) + 64*x1^2*x7^2*(2*x2^2*x7^2 + (x8 - 2*x4*x7)*x10) + 32*x1*x2* x7^2*((x8 - 2*x4*x7)*(x6 - 2*I*x9) + (4*x3*x7 + x5 - 2*I*x9)* x10) + (2*x8 - 4*I*x3*x7 - 4*x4*x7 - I*x5 - 2*x9)*(2*x8 + 4*I*x3*x7 - 4*x4*x7 + I*x5 + 2*x9)*((x6 - 2*I*x9)^2 + 4*x10^2))/(4*(4*x1^2*x7^2 + 8*I*x1*x2*x7^2 - 4*x2^2*x7^2 + (2*x8 + 4*I*x3*x7 - 4*x4*x7 + I*x5 + 2*x9)*(I*x6 + 2*(x9 + x10))))

I want to find all real solutions for $x_1, x_2, \dots, x_{10}$ such that $$ \text{exp} = 0, $$ under the following constraints:

  • All variables $x_1, \dots, x_{10}$ are real.
  • $x_5, x_6, x_7, x_8, x_9, x_{10} > 0$ (strictly positive).
  • $x_1, x_2, x_3, x_4$ can be negative.
  • $x_1$ and $x_2$ cannot be simultaneously zero.
  • $x_3$ and $x_4$ cannot be simultaneously zero.

I tried working with the FindInstance but it doesn't seem to lead to what I'm looking for:

FindInstance[Simplify[exp] == 0, Variables[exp], 1]

({{x1 -> 1, x7 -> Root[{1 + #1^2 & , -1 + 4#1 + 24*#2 - 16*#1*#2 - 32*#2^2 + 16*#2^4 & }, {2, 1}], x2 -> 0, x10 -> 0, x4 -> 1, x8 -> 1, x6 -> 0, x9 -> -1, x5 -> -1, x3 -> 1}}*)

Any suggestion would be greatly appreciated.

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    $\begingroup$ All solutions or one? $\endgroup$ Commented Dec 10 at 16:00
  • 1
    $\begingroup$ Thanks @azerbajdzan, a few solutions would be enough, but if it turns out hard then I would be happy with even one. $\endgroup$ Commented Dec 10 at 16:55

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Here are 42 positive rational solutions all with {x2 -> 0, x3 -> 0} (as this is allowed by OP constraints).

Smallest (regarding numerators and denominators) among them is:

{x1 -> 1, x10 -> 1, x2 -> 0, x3 -> 0, x4 -> 1, x5 -> 48/41, x6 -> 2, x7 -> 1, x8 -> 71/41, x9 -> 1}

Additional three positive integers solutions:

{{x1 -> 3, x10 -> 1, x2 -> 1, x3 -> 2, x4 -> 1, x5 -> 4, x6 -> 2, x7 -> 1, x8 -> 1, x9 -> 3}, {x1 -> 3, x10 -> 1, x2 -> 1, x3 -> 2, x4 -> 2, x5 -> 4, x6 -> 2, x7 -> 1, x8 -> 3, x9 -> 3}, {x1 -> 3, x10 -> 1, x2 -> 1, x3 -> 2, x4 -> 3, x5 -> 4, x6 -> 2, x7 -> 1, x8 -> 5, x9 -> 3}}

Other solutions can be obtained by changing {1, 2, 3} to, say, {1, 2, 5} inside Tuples[{1, 2, 3}, {6}].

exp = (64*x1^4*x7^4 + 64*x2^4*x7^4 + 16*x2^2*x7^2*(4*x3*x7 + x5 - 2*I*x9)*(x6 - 2*I*x9) + 64*x1^2*x7^2*(2*x2^2*x7^2 + (x8 - 2*x4*x7)*x10) + 32*x1*x2* x7^2*((x8 - 2*x4*x7)*(x6 - 2*I*x9) + (4*x3*x7 + x5 - 2*I*x9)* x10) + (2*x8 - 4*I*x3*x7 - 4*x4*x7 - I*x5 - 2*x9)*(2*x8 + 4*I*x3*x7 - 4*x4*x7 + I*x5 + 2*x9)*((x6 - 2*I*x9)^2 + 4*x10^2))/(4*(4*x1^2*x7^2 + 8*I*x1*x2*x7^2 - 4*x2^2*x7^2 + (2*x8 + 4*I*x3*x7 - 4*x4*x7 + I*x5 + 2*x9)*(I*x6 + 2*(x9 + x10)))); eq = exp // Factor // Numerator // ReIm // ComplexExpand; Cases[{Thread[ DeleteCases[{x1, x4, x5, x6, x7, x8, x9, x10}, Alternatives @@ {x5, x8}] -> #], Solve[eq == 0 /. Join[{x2 -> 0, x3 -> 0}, Thread[DeleteCases[{x1, x4, x5, x6, x7, x8, x9, x10}, Alternatives @@ {x5, x8}] -> #]], PositiveRationals]} & /@ Tuples[{1, 2, 3}, {6}], Except[{___, {}}]]; Sort /@ Join @@@ Flatten[Tuples[{{#[[1]]}, #[[2]], {{x2 -> 0, x3 -> 0}}}] & /@ %, 1] exp /. %

enter image description here

{0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}
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  • $\begingroup$ Thanks @azerbajdzan...you are a genius! Could you suggest some tips here--how can one become a pro in Mathematica? $\endgroup$ Commented Dec 10 at 21:39
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    $\begingroup$ @Rob I am not sure how, you should ask some pro ;-) $\endgroup$ Commented Dec 10 at 23:42
  • $\begingroup$ Haha :)). Anyways, thanks a lot for this nice answer. I have a similar question that I posted here:mathematica.stackexchange.com/questions/317245/…. Please take a look. $\endgroup$ Commented 2 days ago

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