Three. (25 points) If the pair of positive integers $(a, x)$ satisfies
$$
\sqrt{\frac{a-x}{1+x}}=\frac{a-x^{2}}{1+x^{2}} \neq x,
$$

find all positive integers $a$ that meet the requirement.
$$
\text { Three, let } \sqrt{\frac{a-x}{1+x}}=\frac{a-x^{2}}{1+x^{2}}=t \text {. }
$$

Then $t^{2} x+x+t^{2}-a=0$,
$$
(t+1) x^{2}+t-a=0 \text {. }
$$
$t \times(2)-x \times(1)$ gives
$t x^{2}-x^{2}-t^{2} x+t^{2}-a t+a x=0$
$\Rightarrow(t-x)(t+x-t x-a)=0$
$\Rightarrow t=x$ (discard) or $t=\frac{a-x}{1-x}$.
Thus $\frac{a-x^{2}}{1+x^{2}}=\frac{a-x}{1-x}$
$\Rightarrow 2 x^{3}-(1+a) x^{2}+(1-a) x=0$.
Given that $x$ is a positive integer,
$$
2 x^{2}-(1+a) x+(1-a)=0 \text {. }
$$

Therefore, the discriminant of equation (1) must be a perfect square. Let $(1+a)^{2}-8(1-a)=k^{2}(k \in \mathbf{N})$, i.e., $(a+5+k)(a+5-k)=32$.
Since $a+5+k$ and $a+5-k$ are both even, and $a+5+k > a+5-k$, we have $(a+5+k, a+5-k)=(16,2)$ or $(8,4)$.
Solving gives $a=4$ or 1.
Substituting into equation (1) yields
$(a, x)=(4,3)$ or $(1,1)$.
Upon verification, only $(a, x)=(1,1)$ satisfies the given conditions, i.e., only $a=1$ meets the requirement.