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1

So the idea is to use the IV of MD5 as a key to create a MAC. Like CodesInChaos mentions in a comment, it would be pretty much equivalent to using $H(k||m)$, if your IV is randomly chosen. By only using it on fixed length messages you avoid the length extension attack, but that is not the only attack on hash constructions that try to create a MAC. In this ...

0

An IV is not a key. In modes of operation, an IV does not have to be kept secret, and in order to decrypt everything you need the IV, which is transmitted in the clear. The only required property of an IV is that it is unique, or at least with overwhelming probability unique. For that you can use "bad" randomness to create it (e.g. from a non-secure PRNG). ...

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In general, no: two bitstrings with the same MD5 need not be related by a simple relation (other than the obvious have the same MD5). Argument: take $2^{128}+1$ distinct random $1024$-bit bitstrings; by the pigeonhole principle, at least two are bound to have the same MD5; since less than $2^{257}$ pairs of distinct bitstrings can be picked among ...

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Here is what I see from the two strings: 0e306561559aa787d00bc6f70bbdfe3404cf03659e704f8534c00ffb659c4c8740cc942feb2da115a3f4155cbb8607497386656d7d1f34a42059d78f5a8dd1ef 0e306561559aa787d00bc6f70bbdfe3404cf03659e744f8534c00ffb659c4c8740cc942feb2da115a3f415dcbb8607497386656d7d1f34a42059d78f5a8dd1ef The bold characters are those that are different between ...

1

Not provably. Maybe, since it's not a permutation, but you couldn't verify it practically. Just use all-zeroes: it's easily recognizable as a null value, and thanks to pre-image resistance, no-one knows what hashes to that.

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