Timeline for Is there an asymmetric algorithm that can perform double encryption?
Current License: CC BY-SA 3.0
9 events
when toggle format | what | by | license | comment | |
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Apr 4, 2017 at 22:13 | history | tweeted | twitter.com/StackCrypto/status/849384429726429187 | ||
Apr 2, 2017 at 18:58 | history | edited | Ilmari Karonen | CC BY-SA 3.0 |
we have mathjax here, we don't need all that extra HTML; also add citation to quote
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Mar 16, 2016 at 12:09 | comment | added | Mok-Kong Shen | I wonder how you got the value 88. If the block size of the plaintext material is mb bits (mb even), It is practically easy to obtain an RSA modulus n (n > 2**mb) that has mb + 2 bits. See Example 2 in Appendix of s13.zetaboards.com/Crypto/topic/7234475/1/ This implies that the differences between the bit sizes of successive elements in the sequence of RSA moduli could be as small as 2. | |
Mar 15, 2016 at 21:30 | history | edited | Maarten Bodewes♦ | CC BY-SA 3.0 |
formatting, title
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Mar 15, 2016 at 14:53 | comment | added | Achille ishak | @Mok-KongShen thanks, I agree with you. i taught i can find other ideas for the same key size. since increasing size is not very suitable and i tested it. must every time increase at least by 88 bit. | |
Mar 15, 2016 at 14:46 | vote | accept | Achille ishak | ||
Mar 15, 2016 at 13:25 | comment | added | Mok-Kong Shen | If you have a sequence of RSA moduli that are strictly increasing and you have plaintext materials seperated into blocks of a constant size that is smaller than the least modulus, you could evidently do that multiple RSA encryption in the ordering of the sequence of the moduli.. | |
Mar 15, 2016 at 9:31 | answer | added | user991 | timeline score: 4 | |
Mar 15, 2016 at 8:52 | history | asked | Achille ishak | CC BY-SA 3.0 |