Given that $a$ is a quadratic residue modulo $p$, we know
$$a^{\frac{p-1}{2}}=a^{2 \cdot 2^{i} u}=\left(a^{u}\right)^{2^{i+1}} \equiv 1(\bmod p)$$

From this, we get $\left(a^{u}\right)^{2^{i}} \equiv +1$ or $-1(\bmod p)$. If $\left(a^{u}\right)^{2^{i}} \equiv -1(\bmod p)$, then $\mu=\lambda$. If $\left(a^{u}\right)^{2^{i}} \equiv 1(\bmod p)$, then we must have $\left(a^{u}\right)^{2^{i-1}} \equiv 1$ or $-1 (\bmod p)$. This implies $\mu=\lambda-1$ or $\left(a^{\mu}\right)^{2^{i-1}} \equiv 1(\bmod p)$, which means $\left(a^{u}\right)^{2^{i-2}} \equiv -1$ or $+1(\bmod p)$. Continuing this process, we eventually get $\left(a^{\mu}\right)^{2} \equiv -1$ or $+1(\bmod p)$. This leads to $\mu=1$ or $a^{u} \equiv \pm 1(\bmod p)$. However, we assumed $a^{u} \not \equiv \pm 1(\bmod p)$, so there must be an integer
$$\mu=1 \text { or } 2 \text { or } \cdots \text { or } \lambda,$$

such that
$$\left(a^{u}\right)^{2^{\prime \prime}} \equiv -1(\bmod p)$$

Among the numbers $1, 2, \cdots, 2^{j}-1$, there are the following $2^{\mu-1}$ numbers:
$$2^{i-\mu}, 2^{i-\mu} \cdot 3, 2^{i-\mu} \cdot 5, \cdots, 2^{i-\mu}\left(2^{\mu}-1\right)$$

If $t$ is odd, we always have
$$\left(\left(b^{2 u}\right)^{2^{i-\mu} \cdot t}\right)^{2^{\mu}}=\left(b^{2 \cdot 2^{\prime} \mu}\right) t \equiv (-1)^{t}=-1(\bmod \cdot p)$$

Thus, $\left(b^{2 u}\right)^{2^{i-\mu}}, \left(b^{2 u}\right)^{2^{i-\mu} \cdot 3}, \left(b^{2 u}\right)^{2^{i-\mu} \cdot 5}, \left(b^{2 u}\right)^{2^{i-\mu} \cdot\left(2^{\mu}-1\right)}$ are $2^{\mu-1}$ solutions to the congruence equation
$$x^{2^{\mu}} \equiv -1(\bmod p)$$

This congruence equation has $2^{\mu}$ solutions, and the other $2^{\mu-1}$ solutions are the negatives of the aforementioned $2^{\mu-1}$ solutions. From the previous discussion, we know that $a^{\mu}$ is a solution to this congruence equation, so there must be an odd number $t$ such that
$$a^{u} \equiv +\left(b^{2 u}\right)^{2^{j-\mu} \cdot t} \text { or } -\left(b^{2 u}\right)^{2^{i-\mu} \cdot t}(\bmod p),$$

which means
$$\left(b^{2 u}\right)^{2^{i-\mu}} \cdot t \equiv +a^{u} \text { or } -a^{u}(\bmod p),$$

Thus, we get $\dot{h}=2^{i-\mu} \cdot t$. Therefore, the proposition is proved.

According to this proposition, when we need to find $h$, we first look for a number in the following set
$$\left(a^{u}\right)^{2}, \left(a^{u}\right)^{2^{2}}, \cdots, \left(a^{u}\right)^{2^{i}}$$

that is congruent to $-1$ modulo $p$. If
$$\left(a^{u}\right)^{2^{A}} \equiv -1(\bmod p)$$

then we must have
$$h=2^{i-\mu} \cdot t$$

Next, we look for a number in the set
$$\left(b^{2 u}\right)^{2^{i-\mu}}, \left(b^{2 u}\right)^{2^{i-\mu} \cdot 3}, \cdots, \left(b^{2 u}\right)^{2^{i-\mu} \cdot\left(2^{\mu}-1\right)},$$

or equivalently, in the set
$$b^{2^{i-\mu+1}} \cdot u, \left(b^{2^{i-\mu+1} \cdot u}\right)^{3}, \cdots, \left(b^{2^{i-\mu+1} \cdot u}\right)^{2^{\mu}-1}$$

that is congruent to $+a^{\mu}$ or $-a^{\mu}$ modulo $p$. This way, we can find the integer $t$, and thus determine $h$.