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In RSA algorithm, the modulus $n$ should be chosen based on the below condition $2^{8(k-1)} \leqslant n < 2^{8k}$ Where $k$ is the byte length of the modulus $n$. for this condition I want the proof with an example. An example is required for the condition: $2^{8(k-1)} \leqslant n$ what is the purpose of this condition? |
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For RSA, we generally count bits rather than bytes; with RSA, we say a modulus is $k$ bits long if $2^{k-1} \le n < 2^k$ There's no proof involved, because it is not a mathematical statement, instead, it is how we define what a $k$ bit modulus is. Another way of looking at it is if we take the modulus, and count the bits, from the highest bit that is set, and including all the bits below that; if there are $k$ bits there, we say that is a $k$ bit number. For example, if we consider the number 187; that has the binary expansion: (where there are a potentially unbounded number of 0's that occur to the left). If we start at the highest 1 bit, and start counting there, we end up with 8 bits; hence we would say that 187 is an 8 bit modulus. This is exactly the same as the first definition, as $2^{8-1} \le 187 < 2^8$ Now, in other contexts, we might use a different definition of the 'number of bits long' something is. For example, on a CPU, we might say that a specific register is '32 bits long'. What that means that the specific register can hold $2^{32}$ distinct settings; we can call the value it holds a 32 bit value (even if the value at the moment happens to be 3). This definition is more about the representation of the integer, rather than the integer itself. We don't use this definition with RSA, because modulii in abstract RSA doesn't have any intrinsic representation. |
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