Cryptography, Information Theory, and Error-Correction. Aiden A. Bruen
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      This indicates that y is 5089 comma StartFraction z Over 55 EndFraction equals 0.745 455 so z equals 55 left-parenthesis 0.745 455 right-parenthesis equals 41.000 025. This is not what we were hoping for: z is supposed to be a whole number, namely the remainder when 279936 is divided by 55! However, the calculator has made rounding errors, and we suspect that z is 41 (and y is 5089). This is easily checked. We can verify that Eq. (3.5) checks out with y equals 5089, z equals 41 since 279 936 equals left-parenthesis 55 right-parenthesis left-parenthesis 5089 right-parenthesis plus 41.

      Principle 1 To calculate the remainder of 279936 when divided by 55, perform the division on a calculator and multiply the decimal part by 55. Verify your answer by checking that Eq. (3.5) is satisfied. This also works to get the remainder whenever we divide a positive integer (= positive whole number) u by another positive integer v.

      Question

      How do we know that z is unique? Maybe there are two possible values?

      Go back to Eq. (3.5), and suppose we have two solutions with y 1 comma y 2 being positive and z 1 and z 2 both lying between 0 and 54. So we have

      (3.8)279 936 equals 55 y 1 plus z 1

      (3.9)279 936 equals 55 y 2 plus z 2

      Then 55 y 1 plus z 1 equals 55 y 2 plus z 2. Now, if y 1 equals y 2 it follows that z 1 equals z 2. So assume that y 1 not-equals y 2. Call the larger one y 1, so y 1 greater-than y 2.

      We now have 55 left-parenthesis y 1 minus y 2 right-parenthesis equals z 2 minus z 1. Since y 1 is at least 1 bigger than y 2, we get that the left side is at least 55. Since z 1 and z 2 are between 0 and 54, we see that the right side is at most 54. Since 55 greater-than 54, we conclude that the assumption y 1 not-equals y 2 leads to a contradiction. Thus, y 1 equals y 2 (and so also z 1 equals z 2): end of story. As a consequence, to check Eq. (3.5) in the future all we need to do in the case above is to ensure that 279 936 minus z is divisible by 55.

      Getting back to our main narrative, A transmits the cipher text bold upper C equals 41 to B having calculated this from the message upper M equals 6. How does B recover upper M from 41? B knows that upper N equals 55 equals left-parenthesis 5 right-parenthesis left-parenthesis 11 right-parenthesis. Since we are using a public cryptosystem, the enciphering algorithm is public knowledge (in this particular example), the enciphering algorithm is “multiply the message by itself seven times and take the remainder on division by upper N”: this gives the cipher text 41. B calculates the deciphering index d as follows.

      There is a unique positive integer d between 1 and 39 such that 7 d gives a remainder of 1 when divided by 40. In this case, it turns out that d equals 23 (more on this later) since left-parenthesis 7 right-parenthesis left-parenthesis 23 right-parenthesis equals 161 and 161 leaves a remainder of 1 on division by 40. Here, 40 comes from the fact that left-parenthesis 11 minus 1 right-parenthesis left-parenthesis 5 minus 1 right-parenthesis equals 40 and 5, 11 are the factors of upper N equals 55.

      To recover the message, B multiplies the cipher text, namely 41, by itself 23 times, gets the remainder on division by 55, and this should give the original message, namely 6. So we are claiming that 4 1 Superscript 23 minus 6 is divisible by 55.

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