COMP 547: Assignment 1 Solutions

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( ) ( ) x x (b) 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 . ( ) ( ) x x 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 . Thus, regardless of how z was chosen, as long as it was chosen correctly, we can see that for each of our categories that an integer x ∈ X can fall into (in both QR p and QR q , QR p but not QR q , QR q but not QR p or neither QR p nor QR q ), either x, yx, zx or zyx is in QR n . Furthermore, there is a one-to-one correspondence between these occurrences. It is quite easy to show that if one of x, yx, zx or zyx is in QR n , then the other three ( possibilities ) ( ) are not in QR ( n ). For ( instance, ) ( ) x x zx z x suppose x ∈ QR n . Then p = q = 1. Either p = p p = ( ) ( ) ( ) zx z x (−1)(1) = −1 or q = q q = (−1)(1) = −1, so zx /∈ QR n . Either ( ) ( ( ) ( ) ( ) ( ) ( ) ( ) zyx z p = y x p) p p = (1)(−1)(1) = 1 or zyx z q = y x q q q = ) ( ) ( ) (1)(−1)(1) = 1, so zyx /∈ QR n . = y p = (−1)(1) = −1, so ( yx p yx /∈ QR p . The cases of yx, zx and zyx being in QR n can all be dealt with in a similar fashion. We have shown that if n, y and z were correctly chosen, we should obtain results similar to those we did, in fact, obtain: a somewhat even categorisation of the 100 integers in X into our four type categories. However, does the converse hold That is, if a more or less even categorisation of all the integers in X occurred, can we be certain that n, y and z were correctly chosen First, we note that it is easy to verify that ( ( y n) = 1 and z n) = −1. This can be done by simply computing the Jacobi symbols in question. Therefore, in the rest of this discussion, we will assume that ( y ) n) = 1 and = −1. ( z n Now, suppose n has more than two prime factors, that is, n = p 1 · · · p k , where k > 2. x ∈ QR n if and only if x ∈ QR p1 , . . . , x ∈ QR pk . Similarly, yx ∈ QR n if and only if yx ∈ QR p1 , . . . , yx ∈ QR pk , zx ∈ QR n if and only if zx ∈ QR p1 , . . . , zx ∈ QR pk and zyx ∈ QR n if and only if zyx ∈ QR p1 , . . . , zyx ∈ QR pk . Intuitively, it is quite clear that in this case, there will be some integer x ∈ X such that neither x, yx, zx nor zyx is in QR n . x p 28
There are simply too many conditions to be satisfied, as a result of the many prime factors of n. To show this, we must simply find an x ∈ X with, for instance, the following properties: ( ) ( ) • For some p i , y x p i p i = −1. ( ) ( ) z x • For some p j , p j p j = −1. ( ) ( ) ( ) z • For some p k , y x p k p k p k = −1. ( ) x • For some p l , p l = −1. It is clear that this will guarantee that x /∈ QR n , yx /∈ QR n , zx /∈ QR n and zyx /∈ QR n . Note that we could simply try to construct such an x. We begin by choosing a p i to clash with y, then a p j to clash with z, and so on. Of course, this raises the question of whether or not an integer x with the required properties even exists in X. Here, we will simply use intuition and argue that in the case that n = pq, our categoration was tight and complete, in the sense that every quadratic residue combination an x could fall into corresponded exactly to one of our QR n types (x, yx, zx or zyx). Here, we have far more quadratic residue combinations, since n has more than two prime factors. Four of these combinations may still correspond to our x, yx, zx and zyx types, as described above. The other combinations, however, might correspond to situations where neither x, yx, zx nor zyx is in QR n . Therefore, if n = p 1 · · · p k , where k > 2, then summing up the numbers of the variants of x in QR n would yield less than 100. Henceforth, let us assume again that n = pq. Now, we must address the question of whether or not y and z were correctly chosen. As discussed previously, it is easy to verify that ( ( y n) = 1 and z n) = −1. From this, it immediately follows that z is not in QR n , since n has( only ) two ( ) prime factors. The only incorrect case we must consider is that y p = y q = 1, in which case y would be in QR n . In( this ) case, ( ) consider an x ∈ X that is not in QR p and not in QR q . That is, = = −1. Clearly, x /∈ QR n . ( ) ( ) ( ) ( ) ( ) ( ) Also, it follows immediately that yx p = yx q = y x p p = y x q q = (1)(−1) = −1. Thus, yx /∈ QR n . Now, we must again consider our two cases (a) and (b) for z: ( ) ( ) ( ) ( ) ( ) z z zx z x (a) If p = 1 and q = −1, then p = p p = (1)(−1) = −1, ( ) ( ( ) ( ) so zx /∈ QR n . Also, zyx z p = y x p) p p = (1)(1)(−1) = −1, so x p x q zyx /∈ QR n 29
Page 1 and 2: COMP 547: Assignment 1 Solutions Oc
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COMP 547: Assignment 1 Solutions

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