I know that in Bitcoin the difficulty target adjust after creating 2016 blocks. So, it means that within these 2016 block the difficulty target remains same (static). Am I correct?
If this is correct then the miners has to find the right Hash of the header. Header=version+hash of previous blockheader+merkle tree root+timestamp+difficulty target+nonce variable.
Now, it seems that within these 2016 blocks the input (block header) to the hash function will be different but the output (difficulty target) remains the same. Therefore it seems to me that it breaks the property of "strong collision resistance" property of the hash. Am I correct? In other word, Bitcoin uses this property of the hash function to make it difficult to find solution of the puzzle.
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