Deriving least squares coefficients for curve of form $y=a/(x^2+b)$ I am trying to derive the coefficients $a$ and $b$ for a least squares curve of the form $y=\frac{a}{x^2+b}$
To start I have written $$S=\sum_i(y_i-y)^2=\sum_i\left(y_i-\frac{a}{x_i^2+b}\right)^2=\sum_i\left(y_i^2-\frac{2a y_i}{x_i^2+b}+\frac{a^2}{(x_i^2+b)^2}\right) \\S=\sum_iy_i^2-2a\sum_i\frac{y_i}{x_i^2+b}+a^2\sum_i\frac{1}{(x_i^2+b)^2}$$
Now the way I have seen it done for the line $y=a+bx$ is they minimize $S$ with respect to both $a$ and $b$, applying that for this curve gives the two constraint equations:
$$0=a\sum_i\frac{1}{(x_i^2+b)^2}-\sum_i\frac{y_i}{x_i^2+b}\\
0=-a\sum_i\frac{1}{(x_i^2+b)^3}+\sum_i\frac{y_i}{x_i^2+b}$$
Summing these together we find $$0=a\left(\sum_i\frac{1}{(x_i^2+b)^2}-\sum_i\frac{1}{(x_i^2+b)^3}\right)$$
Since $a\neq 0$ we must have $$\sum_i\frac{1}{(x_i^2+b)^2}=\sum_i\frac{1}{(x_i^2+b)^3}$$ Which seems to imply $$(x_i^2+b)^2=(x_i^2+b)^3$$ and unfortunately there doesnt appear to be a constant $b$ that solves this equation. Did I make a mistake somewhere or is this the result?

We have $$\frac{\partial S}{\partial a} = -2\sum \frac{y_i}{x_i^2+b}+2a\sum \frac{1}{(x_i^2+b)^2}$$ $$\frac{\partial S}{\partial b} = 2a \sum \frac{y_i}{(x_i^2+b)^2} -2a^2 \sum \frac{1}{(x_i^2+b)^3}$$
