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Why is there no solution for the system of equations with a greater than 0?

Solve[{c^2 - b^4/a^2 == 0, a^2 + b^2 == c^2, e == c/a, a > 0, 
  e > 1}, e, {a, c}]

Why can we solve the value of e by setting a not equal to 0?

Solve[{c^2 - b^4/a^2 == 0, a^2 + b^2 == c^2, e == c/a, a != 0, 
  e > 1}, e, {a, c}]

{{e -> 1/2 (1 + Sqrt[5])}}

How can we solve this equation if a is greater than 0?

Solve[{(((2 Sqrt[2] + 2) a)^2 + 8 a^2 - 4 c^2)/(
   2 (2 Sqrt[2] + 2) a 2 Sqrt[2] a) == Sqrt[2]/2, e == c/a, a > 0, 
  e > 1}, e, {a, c}]

the result is:

{{e -> Sqrt[3]}}
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    $\begingroup$ Solve[{c^2 - b^4/a^2 == 0, a^2 + b^2 == c^2, e == c/a, a > 0, e > 1}, e, {a, c}, Reals] $\endgroup$
    – cvgmt
    Mar 12 at 2:54
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    $\begingroup$ The result of Solve[{c^2 - b^4/a^2 == 0, a^2 + b^2 == c^2, e == c/a, a > 0, e > 1}, e, {a, b}, Reals] is more informative. $\endgroup$
    – user64494
    Mar 12 at 6:59
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    $\begingroup$ Looks like a bug to me. I would report it to "[email protected]" $\endgroup$
    – Daniel Huber
    Mar 12 at 9:02

2 Answers 2

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There are no solutions with generic complex b -- Im[b] needs to be zero.

In[1]:= Solve[{c^2 - b^4/a^2 == 0, a^2 + b^2 == c^2, e == c/a, a > 0, e > 1}, e, {a,
 c}, MaxExtraConditions->1]//InputForm                                              

Out[1]//InputForm= 
{{e -> ConditionalExpression[(1 + Sqrt[5])/2, (Im[b] == 0 && Re[b] > 0) || 
     (Im[b] == 0 && Re[b] < 0)]}}

There is a solution with generic real b.

In[2]:= Solve[{c^2 - b^4/a^2 == 0, a^2 + b^2 == c^2, e == c/a, a > 0, e > 1}, e, {a,
 c}, Reals]//InputForm                                                              

Out[2]//InputForm= {{e -> (1 + Sqrt[5])/2}}
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  • $\begingroup$ Solve[{c^2 - b^4/a^2 == 0, a^2 + b^2 == c^2, e == c/a, a != 0, e > 1}, e, {a, b, c}] $\endgroup$
    – csn899
    Mar 13 at 5:57
  • $\begingroup$ Solve[{c^2 - b^4/a^2 == 0, a^2 + b^2 == c^2, e == c/a, a > 0, e > 1}, e, {a, b, c}] $\endgroup$
    – csn899
    Mar 13 at 5:57
  • $\begingroup$ After your explanation, we know that the solution of this equation is related to whether the parameter b is a complex number whose imaginary part is not equal to 0. The value of e can be calculated successfully in the above two forms. {a,b,c} $\endgroup$
    – csn899
    Mar 13 at 6:00
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I opine that your original problem was poorly stated.

I took the liberty of restating yor problem after noting the $e$ could computed afterwards:

solution = Reduce[{c^2 - b^4/a^2 == 0, a^2 + b^2 == c^2, 
c/a > 0, a > 0}, {a, b, c}, Reals]

I used Reduce to attempt to avail all the solutions.

ConditionalExpression[{ToRules[solution[[2]]]}, a > 0]

$$\fbox{$\left\{\left\{b\to -\sqrt{\frac{1}{2} \sqrt{5} \sqrt{a^4}+\frac{a^2}{2}},c\to \frac{b^2}{a}\right\}, \\ \left\{b\to \sqrt{\frac{1}{2} \sqrt{5} \sqrt{a^4}+\frac{a^2}{2}},c\to \frac{b^2}{a}\right\}\right\}\text{ if }a>0$}$$

ConditionalExpression[{{b -> -Sqrt[a^2/2 + (1/2)*Sqrt[5]*Sqrt[a^4]], c -> b^2/a}, {b -> Sqrt[a^2/2 + (1/2)*Sqrt[5]*Sqrt[a^4]], c -> b^2/a}}, a > 0]

Of course, I may have missed something.

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