title: Hash collision H(x)=H(y) with x!=y, does that also mean H(x+z)=H(y+z) for random z?
body: <p>Suppose H is some hash function (such as MD5 or SHA256 or whatever) and I have a collision for this hash: two different pieces of data x and y, that have the same hash.</p>
  
  <p>In other words x≠y but H(x)=H(y).</p>
  
  <p>Now if I concatenate some random data z, will H(x+z) be the same as H(y+z) ?</p>
  
  <p>The idea is: x and y being a collision may imply that they happen to bring the H function in the same state (thus resulting in the same hash). From that point on, it doesn't matter what other data we append, their hashes will remain equal.</p>
  
  <p>I tested the above for <a href="http://www.mscs.dal.ca/~selinger/md5collision/" rel="nofollow">this MD5 collision</a> and it seemed to work there. But I don't know if this is true in general?</p>