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Maybe OP just xor'd the files with each other and the result was the RANDOM_STRING string followed by a pile of \0 bytes, repeat.

Edit: To clarify, I meant the encrypted and temporarily-decrypted-by-the-wrapper-script versions of the same .mov file.

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The XOR of two ciphertexts under P XOR K isn't K (you have to know a plaintext to make that trick work, which, granted, you likely would here). The solution that doesn't require a known plaintext is comparably simple.

As the post says, he gained access to both decrypted and encrypted version. Then (p XOR k) XOR p = k.

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