Tag Info

New answers tagged

1

Since $p$ and $q$ are primes, the only factors you needs to rule out are those two numbers. Suppose $p$ divides $(p-1)(q-1)$. Then it divides either $p-1$ (clearly not true) or $q-1$. The latter means $q-1 = p \cdot x$, for some $x \ge 2$ (if $x = 1$ either $p$ or $q$ is even, which is only possible if the numbers are 2 and 3). However, then $q \ge 2p+1$, ...


5

The only reason you are seeing this is because you are dealing with such small primes. With primes like we would use in practice (1024 bits), the probability of this happening is very, very small. And, it can only happen when $e>\sqrt{\lambda(n)}$. Since we typically use $e=65537$ in practice, it is guaranteed to not happen. Anyways, there is no mistake ...


6

The table from the Handbook of Applied Cryptography is computed for candidate primes taken as odd integers. This is equivalent to using trial division by $2$ only. Note that section 4.48 specifies that: Using more advanced techniques, the upper bounds on $p_{k,t}$ given by Fact 4.48 have been improved. These upper bounds arise from complicated formulae ...


7

Update: the formula actually on page 165 in chapter 4 of the HAC is: $t=\lceil{1\over2}\cdot\lg n\rceil$, where $\lg$ is the base-2 logarithm; that is $t=\lceil{1\over2}\cdot k\rceil$; and that crude estimate asks for 512 Miller-Rabin tests for 1024-bit primes, not 5 or 6 tests as stated in the question based on the erroneous $t=\lceil{1\over2}\cdot\lg ...



Top 50 recent answers are included