COMP 547: Assignment 1 Solutions

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Type x yx zx zyx Amount 24 23 27 26 Table 1: Summary of results for problem 11. 5) 5471234567988 is not in QR_n (root of yx: 125637127136393030\ 900524755578585052721630258447382451765030144949951207786646\ 110838585885971204218719872431155751276782292634789749171217\ 050399637909286063835079790158907905880286681588957215994637\ 4) 5471234567989 is not in QR_n (root of yx: 130096871661902070\ 710031752707137582830741110348039577107913444950334782429346\ 834499138925114821873522260732875624204023958322262242726896\ 915774023449924304056024312035505160809645729547061229283948\ 0) 24, 23, 27, 26 These results are summarised in table 1. 12. We will first propose a way to check that n, y and z have all been chosen correctly and then discuss how and why it works. Let X be a set of 100 random number in Z ∗ n. Here is a way to check that n, y and z have all been chosen correctly: (a) Compute ( y n) . If it is not 1, return fail. (b) Compute ( z n) . If it is not −1, return fail. (c) Sum up the amounts of variants of x ∈ X that are in QR n (that is, the numbers of x’s, yx’s, zx’s and zyx’s in QR n ). If they do not sum up to 100, return fail. (d) If they do sum up to 100, they should be more or less evenly distributed in the four categories. At this point, it is safe to return succeed. Let us assume, for the moment, that n, y and z have all been chosen correctly. We will make some important observations concerning our results in this case of correctly chosen parameters. We know that given a prime r, exactly half of the integers between 1 and r − 1 are in QR r . Thus, using some intuition, if n = pq, we would expect about half of the integers in X to be in QR p and about half of them to be in QR q . Of course, it will generally not be the case that exactly half of the 26
integers in X fall into either of these two categories, but since |X| = 100 and thus covers a fairly large range, it is reasonable to expect such an approximate distribution. If for some x ∈ X ( y n) is equal to zero, then we can just select a new point at random to replace x in X. Taking this a step further and combining QR p and QR q , we would expect about one fourth of the integers in X to be in neither QR p nor QR q , about one fourth to be in QR p but not QR q , about one fourth to be in QR q but not QR p and about one fourth to be in both QR p and QR q . Again, the distribution might be very skewed and need not be even. However, we would expect there to be a number of integers in X that fall into each of these four categories. This accounts for all 100 integers in X. The important observation we will make is that these four cases (x being in QR p and QR q , QR p but not QR q , QR q but not QR p or neither QR p nor QR q ) correspond to the cases of x, yx, zx or zyx being in QR n . Let us explore this idea. ( ) ( ) ( ) ( ) We know that y p = y z z q = −1 and that either (a) p = 1 and q = ( ) ( ) −1 or (b) = −1 and = 1. (Recall that, for now, we are assuming z p that n, y and z were chosen correctly.) z q Now, consider some x ∈ X. ( ) ( ) x x • If x ∈ QR p and x ∈ QR q , then p = 1 and q = 1. From this, we can immediately conclude that x ∈ QR n . ( ) ( ) x x • If x /∈ QR p and x /∈ QR q , then p = −1 and q = −1. Now, ( ) ( ) ( ) using y, we can see that yx p = y x p p = (−1)(−1) = 1 and ( ) ( ) ( ) yx q = y x q q = (−1)(−1) = 1. From this, we can conclude that yx ∈ QR n . ( ) z • Next, we must consider our cases for z. Recall that either (a) p ( ) ( ) ( ) = 1 z z z and q = −1 or (b) p = −1 and q = 1. ( ) ( ) x x (a) If x ∈ QR p and x /∈ QR q , then p = 1 and q ( ) ( ) = −1. Using ( ) z z zx the facts that p = 1 and q = −1, we see that p ( ) ( ) ( ) ( ) ( ) = z x zx z x p p = (1)(1) = 1 and q = q q = (−1)(−1) = 1. From this, we can conclude that( zx) ∈ QR n . ( ) x x If x /∈ QR p and x ∈ QR q , then p = −1 and q = 1. Using ( ) ( ) ( ) ( ) z z the facts that p = 1, q = −1 and y p = y q = −1, ( ) ( ( ) ( ) we see that zyx z p = y x p) p p = (1)(−1)(−1) = 1 and ( ) ( ) ( ) ( ) zyx z q = y x q q q = (−1)(−1)(1) = 1. From this, we can conclude that zyx ∈ QR n . 27
Page 1 and 2: COMP 547: Assignment 1 Solutions Oc
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Page 21 and 22: 5471234567945 is in QR_n (root of x
Page 23 and 24: 5471234567962 is not in QR_n (root
Page 25: 5471234567979 is not in QR_n (root
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COMP 547: Assignment 1 Solutions

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