# Why is it possible to solve this equation by setting a greater than 0 and not equal to 0?

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)}}


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

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


the result is:

{{e -> Sqrt}}

• 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] Mar 12 at 2:54
• 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. Mar 12 at 6:59
• Looks like a bug to me. I would report it to "[email protected]" Mar 12 at 9:02

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

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


There is a solution with generic real b.

In:= 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//InputForm= {{e -> (1 + Sqrt)/2}}

• 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}] Mar 13 at 5:57
• 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}] Mar 13 at 5:57
• 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} Mar 13 at 6:00

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


Of course, I may have missed something.