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Mar 25, 2022 at 17:02 vote accept J. Doe
Mar 25, 2022 at 14:56 comment added J. Doe ah ok, '$(2Fpq+p+q)$' may be factorized easily. How about we take care about this and set it to a small number times a $500$-bit prime? Would it be still easy?
Mar 25, 2022 at 14:46 comment added J. Doe is factoring '$N-1 = 2F(2Fpq+p+q)$' that easy? It's still an 922-bit number. How much easier is it compared to a regular used 922-bit number? Current record of a hard number is 829-bit. If it is easy. Does it significantly depend at the size of $F$? We could scale it as big as we want as long $p,q$ have a constant size.
Mar 25, 2022 at 14:17 history answered MostlyResults CC BY-SA 4.0