1.Is there a integer x(express in bits...), so that.
md5(x) = x
2.If there is, how many are there.
3.If there aren't any, what's the least amount of md5 need to be applied to any x resulting itself.

There are 2^128 possible x to consider. Yes, Beyond our computation power

Suppose that md5(x) is a function that randomly maps an x with another non-negative integer that's smaller than 2^128(not necessarily one to one). There are some expected values.

Expected number of x with that property out of all possible x with above assumption.
I would think it as pick x, then randomly pick a number from 0 to 2^128-1 for f(x). Each pick are mutually exclusive. What's the chance x = f(x)
There are 2^256 combination of (x, f(x)), 2^128 of them are x = f(x)
Then we have...... 1 out of 2^128.
Doesn't give us too much hope... haha... This problem is not likely to be answered brutal force unless I'm really lucky or md5 is designed with like 10^20 md5(x) = x.

I'm searching for an analytical method to do No.1
2 and 3 are likely only brutal forceable. While there isn't enough computational power to do it.

This problem can be generalized into any hash function that's complex(beyond normal humans' analytical thinking) and large(beyond any computer's computational power) to show how limited our knowledge are.

My site have been gaining people with strange fetishes.