A different type of algebra problem 2

It's an really interesting problem & I'm just posting my approach of solution as I'm unsure about it.

Expanding we get ,

$x+x^4+x^9+x^{16}+x^{25}=\frac{1}{x} + x^4 + \frac{1}{x^{27}} + x^{256} + \frac{1}{x^{3125}}$

So Just transforming it into a equation by multiplying $x^{3125}$ to both sides we have ,

$f(x)=x^{3381} - x^{3150} - x^{3141} - x^{3134} - x^{3126} + x^{3124} + x^{3098} + 1=0$

$f(-x)=-x^{3381} - x^{3150} + x^{3141} - x^{3134} - x^{3126} + x^{3124} + x^{3098} + 1=0$

Applying Descarte's Rule of sign to the polynomial we have there are two sign changes in $f(x)$ i.e.

$\color{#D61F06}+x^{3381}$ to $\color{#3D99F6}-x^{3150}$ & $\color{#3D99F6}-x^{3126}$ to $\color{#D61F06}+x^{3124}$

So it has atmost 2 positive real roots.

Again we have three sign changes in f(-x) i.e. ,

$\color{#3D99F6}-x^{3150}$ to $\color{#D61F06}+x^{3141}$ & $\color{#D61F06}+x^{3141}$ to $\color{#3D99F6}-x^{3134}$ & $\color{#3D99F6}-x^{3126}$ to $\color{#D61F06}+x^{3124}$ .

So it has atmost 3 negetive real roots .

Now in $f(-x)$ we have Product of roots = $-\frac{a_n}{a_0} =$ $-\frac{1}{-1}=1$ So there are 2 or 0 negetive roots .

In the original equation we have product of roots = $-1$ so we must have a negative root. The rest of the roots are complex conjugates whose product is always positive as $z.\bar{z}=(|z|)^2$

So we have $\boxed{1}+\boxed{2}=\boxed{3}$ real solutions.