Sorry this explanation is kind of poorly written.

Note that for x coprime to 2^32, x^(2^30) = 1 mod 2^32.

Suppose that a has maximum order, i.e. a^k ≠ 1 for 0 < k < 2^30.

Let n = c0 + c1*2 + c2*2^2 + ... + c29*2^29.

(a^n)^(2^29) = c^(2^29)

so

a^(c0 * 2^29) = c^(2^29)

Since a has maximum order, exactly one of c0 = 0 and c0 = 1 will work. And then you can proceed like that to find the other ci.

So where does LTE come in?

Well, since a and c are both 1 mod 4, v2((a^c0)^(2^29) - c^(2^29)) = v2(a^c0 - c) + v2(2^29).

We know the LHS is either 32 or <32 depending on choice of c0. So RHS is either 3 or <3 depending on choice of c0.

Similar argument applies for other ci.

Why v2(a-1) = v2(c-1)?

Well, earlier I had this assumption that a had maximal order. That may not be true. However, if a = 1 mod 4, then o(a) = 32 - v2(a - 1).

So a and c have the same order.

Let that order be 2^h.

Then we can do stuff like

a^(c0*2^(h-1)) = c^(2^(h-1)) for exactly one choice of c0.