Can the solution to this equation be written as the root of a polynomial?

Question

While thinking about a question on another Stack, I ended up trying to solve the following equation for $a$: $$ \sqrt{\frac{2}{a} - \frac{2}{a+1}} - \sqrt{\frac{1}{a}} + 1 - \sqrt{2 - \frac{2}{a+1}} = \sqrt{2} - 1 $$ Numerically, it seems that there is a solution at $a \approx 0.490981...$. However, neither I nor Mathematica was able to get this equation into a form where it is obviously an algebraic number.

Can the above solution be written as the root of some polynomial? If so, which one? If not, why not? (It would be surprising to me that what looks like an algebraic equation, involving square roots and simple rational functions, does not have an solution in the algebraic numbers, even for some polynomial of high degree.)

(For those who are curious: $a$ corresponds to the radius, in AU, for which it is more costly to get into a circular orbit at that radius than it is to leave the solar System.)

Answer 1

It is a root of the quartic equation:

2048 sqrt(2) x^4 - 3072 x^4 + 6656 sqrt(2) x^3 - 8192 x^3 + 3072 sqrt(2) x^2 - 6208 x^2 - 1536 sqrt(2) x + 2944 x - 64 = 0

Nicely formatted:

$$ 2048 \sqrt2 x^4 - 3072 x^4 + 6656 \sqrt2 x^3 - 8192 x^3 + 3072 \sqrt2 x^2 - 6208 x^2 - 1536 \sqrt2 x + 2944 x - 64 = 0 $$

Synopsis:

Multiply both sides by $\sqrt{a(a+1)}$.

Use the substitution $y = a + 1/2$

Then rearrange the equation such that only one side contains radicals of the variable $y$. Then square both sides.

Repeat the step above.

Back substitute and simplify.

Let me know if you need more details.

Answer 2

Using Maple's Groebner Basis package, we find that $a$ satisfies the equation $$ 256a^8-1024a^7-2080a^6+3520a^5+3105a^4-4060a^3+1158a^2-92a+1=0 $$ Here's the Maple code . . .