Challenge "Hybrid Encryption I" ¶
By: admin on Nov. 2, 2010, 6:47 p.m.
Hybrid encryption is widely used in practice. However, when using hybridencryption you should take care of certain details…
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By: admin on Nov. 2, 2010, 6:47 p.m.
Hybrid encryption is widely used in practice. However, when using hybridencryption you should take care of certain details…
Read more...
By: Schoetti on Nov. 4, 2010, 1:52 p.m.
Please be aware that you get the wanted solution for this challenge only if you interpret the ciphertext explicit without padding!
Even if you interpret the given ciphertext as if the last byte was padding you will get a meaningful solution but it is not the same as without padding!
Cheers
Pascal
By: wowmyst on Jan. 20, 2011, 3:59 p.m.
really nice, and boy am i proud now for having some luck in solving it. i understand that the main weakness of this specific cipher is the small value for e.
but would even a slightly bigger e=5 be much harder to crack with the same n and a corresponding c_p=0x41fdf1bc054a7fa7920f117c9f34deb404dc7093109b4f07fa0955c1f7ab7f
0c04d7560f94a81ab818e5b7d48d9afbdd0f26c13541fc524bf815e60d961504fc ?
thanks
/edit by admin: added a linebreak to keep the correct width in this forum
By: Schoetti on Jan. 22, 2011, 9:55 p.m.
Hello Wolfgang,
thanks for the feedback!
The challenge is, in principle, quite easy if you know how to solve it. I cannot give you the exact reason why this easy solution would only work with e=3 without giving too much information about the solution. But if you are interested you can ask me in a PM and I will explain this matter to you. When I am informed right, there is another challenge planned with an higher e, but I do not know when this challenge will be online.
Cheers
Pascal
By: Veselovský on Aug. 9, 2011, 1:12 a.m.
Hi,
can anybody give some hints how to use the fact that e=3 (i.e. is small enough) to solve the problem?
I have taken 3rd root of c, but it is not an integer…
By: fretty on Aug. 9, 2011, 10:49 a.m.
Right, so I have converted N into decimal and taken its cube root to get a 51 digit number.
We know that the plaintext for the RSA part is a 128-bit AES key but even the biggest possible key (ffffffffffffffffffffffffffffffff) has 39 digits in decimal form, meaning that the cube of any AES key will be less than N.
Yet the RSA ciphertext has no integer cube root!
Am I doing something wrong?