Given the positive odd numbers $a, b, c, d$ such that $ac - b > 0$ and $(d - a)^2 > (c - b)^2$. From these and the condition $ad = bc$, it follows that $(d + a)^2 > (c + b)^2$, which implies
$$a + d > b + c.$$

From this and equation (2), condition (1) leads to the important numerical relationship
$$k > m \geqslant 3.$$
(ii) From condition (1) and equation (3), we have
$$a(2^k - a) = b(2^m - b), \quad b^2 - a^2 = 2^m b - 2^k a = 2^m \left(b - a \cdot 2^{k-m}\right),$$
where $2^{k-m}$ is a positive integer power of 2. This allows us to use divisibility properties for further discussion.

Since $a, b$ are odd, $(b + a) / 2$ and $(b - a) / 2$ are both integers. Therefore, the second equation in (4) can be written as
$$(b + a) / 2 \cdot (b - a) / 2 = 2^{m-2} \left(b - a \cdot 2^{k-m}\right).$$

Given $m \geqslant 3$, $2^{m-2}$ is a positive integer power of 2. Additionally, $2^{k-m}$ is also a positive integer power of 2. Since $(b + a) / 2$ and $(b - a) / 2$ sum to $a$ (an odd number), one of them must be odd and the other even. According to Chapter 1, §2, Example 2 (ii), $2^{m-2}$ must divide the even one. However, determining the parity of $(b + a) / 2$ and $(b - a) / 2$ is not straightforward, which is another challenge in this problem.

Noting that one of $(b + a) / 2$ and $(b - a) / 2$ is divisible by $2^{m-2}$, and a positive divisor $t$ of a positive integer $r$ is always less than or equal to $r$ (see Chapter 1, §2, Theorem 1(vi)), if we can prove
$$(b - a) / 2 < 2^{m-2},$$

then it follows that $2^{m-2}$ must divide $(b + a) / 2$. This requires using the size relationships discussed earlier. $\square$
(iii) Proof of equation (6). From the conditions $a < b < c < d$ and $b + c = 2^m$, we have
$$(b - a) / 2 < b / 2 < (b + c) / 4 = 2^{m-2}.$$
(iv) From the conditions $b + a < b + c = 2^m$ and $2^{m-2}$ divides $(b + a) / 2$, it follows that
$$2^{m-1} = b + a,$$

and using equation (3) and the above equation, we get
$$(b - a) / 2 = \left(b - a \cdot 2^{k-m}\right), \quad b = \left(2^{k-m+1} - 1\right) a, \quad a = 2^{2m - k - 2}.$$

Since $a$ is odd, we have
$$2m - k - 2 = 0, \quad a = 1, \quad k = 2m - 2.$$

Additionally, we obtain
$$b = 2^{m-1} - 1, \quad c = 2^{m-1} + 1, \quad d = 2^{2m-2} - 1.$$

Proof complete.