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3.3.4 ANSWERS TO EXERCISES 543 3. a2 E 2a -1 and a3 = 3a -2 (modulo m). By considering all short solutions of (15), we find that V: = 6 and V: = 4, for the respective vectors (1, -2,1) and (1, -1, -1, l), except in the following cases: m = 2 q, q odd, e 2 3, a E 2e-1 (modulo 2 ), a E 1 (modulo q), vi = vi = 2; m = 3 q, gcd(3, q) = 1, e > 2, a = 1 & 3e- (modulo 3 ) a = 1 (modulo q), vi = 2; m = 9, a = 4 or 7, V; =Yi = 5. 4. (a) The unique choice for (~1, ~2) is &((Y~UZZ -~2~21, -ylu12 + ysuil), and this is = &(y1u22 + y2au22, –ylul~ -yzauiz) = (0,O) (modulo 1); i.e., ~1 and 52 are integers. (b) When (x1,52) # (O,O), we have (ziuri + ~2usi)~ f (51~2 $ ~2~22)~ = z~(u:, + ut2) + z~(u& + u&) + ~Z~ZZ(U~~UZ~ + ur2u22), and by hypothesis this is L (d +x2 -IW21)(~11 + G2) 2 $1 + 42. [Note that this is a stronger result than Lemma A, which tells us only that Z; 2 (UT, + uf2)(u& + ui2)/m2 and that d I (& +~U:2)~/m , where the latter can be 2 1. The idea is essentially Gauss s notion of a reduced binary quadratic form, Disq. Arith. (Leipzig: 1801) li l.] 5. Conditions (30) remain invariant; hence h cannot be zero in step S2, when a is relatively prime to m. Since h always decreases in that step, S2 eventually terminates with u2 + w2 2 s. Note that pp 5 0 throughout the calculation. The hinted inequality surely holds the first time step S2 is performed. The integer g that minimizes (h -qh)2 + (p -q p)2 is q = round((h h + p p)/(h2 + p )), by (24). If (h -q h)2 + (p -q p)2 < h2 + p2 we must have q # 0, q # -1, hence (p -q p)2 2 p2, hence (h -q h)2 < h2, i.e., Ih -q hJ < h, i.e., q is q or q + 1. We have hu + pv 2 h(h -q h) + p(p -q p) 2 -$(h2 f p ), so if u2 + v2 < s the next iteration of step S2 will preserve the assumption in the hint. If u2 + v2 2 s > (u -h)2 + (v -p) , we have 2/h(u -h) + p(v -p)l = 2(h(h -u) + p(p -v)) = (u -h)2 + (v -p) + h2 + p2 -(u + v ) 2 (u -h)2 + (v -P)~ 5 h2 + p2, hence (u-h)2+(v-p)2 is minimal by exercise 4. Finally if both u2+v2 and (u-h)2+(v-p)2 are 2 s, let u = h -q h, II = p -q p; then 2lhu + pv l 5 h2 + p2 5 d2 + d2, and h2 + p2 is minimal by exercise 4. 6. If u2 +v2 2 s > (u-h)2 +(v-p) in the previous answer, we have (w–P)~ > v2, hence (u -h)2 < u2; and if q = u3, so that h = u,h + u, we must have a,+1 = 1. It follows that ~2 = minc~3 2mjp,-1, hence 7. We shall prove, using condition (19), that u, . uk = 0 for all k # j iff Vj . Vk = 0 for all k # j. Assume that U, uk = 0 for all k # j, and let U, = CYIVI $- . . . + at&. Then U, . vk = ok for all k, hence U, = cyJV3, and V; . fi = cxT1(U3 . fi) = 0 for all k # j. A symmetric argument proves the converse. 8. Clearly vt+l < vt (a fact used implicitly in Algorithm S, since s is not changed when t increases). For t = 2 this is equivalent to (mp2/.rr) /2 > ($mps/.rr)1/3, i.e., P3 L 4&%2 3/2 . This reduces to $10e4 /fi with the given parameters, but for large m and fixed ~2 the bound (39) is better. 9. Let f(yl, . . . , yt) = 8; then gcd(yl, , yt) = 1, so there is an integer matrix W of determinant 1 having (2~1, . . , Wt) as its first row. (Prove the latter fact by induction on the magnitude of the smallest nonzero entry in the row.) Now if X = (21,. . . , xt) is a row vector, we have XW = X iff X = X W- , and W-l is an integer matrix of determinant 1, hence the form g defined by WU satisfies g(zl, . . . , xt) = f(zi, . . . , 2:); furthermore g(l, 0,. . . ,O) = 0.
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