Actually, each unique value of the 8-byte checksum represents 1.9212776604857023e+24 different 4088 byte long block values. So if the 8-byte "hash code" matches then you have 1 of the 1.9212776604857023e+24 4088 byte blocks that will generate that code. However if the 8-byte recorded hash *does not* match the 4088 byte block from which it was generated, you can be absolutely certain that the block of 4088 bytes is not **any** of the 1.9212776604857023e+24 sequences that would generate that value. **NB** this is assuming an "equal distribution" -- the better the distribution, the better (or more secure, to use the accepted parlance) the generated hash. That is why SHA2-512 is better than a parity byte -- simply because it is longer.