By Ionin Y. J., Shrikhande M. S.
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Extra resources for (2s1) Designs withs intersection numbers
4 PROBLEM S 1. Show that if ordp a = i, ordp b = u, and (t, u) = 1, then ordp (ab) = tu. 2. Show that if p = 1 (mod 4) and ^ is a primitive root of p, then so is —g. Show by a numerical example that this need not be the case if p = 3 (mod 4). 3. Show that if p is of the form 2m+ 1 and (a/p) = —1, then a is a primitive root of p. 4. Show that if p is an odd prime and ordp a = t > 1, then t- i —1 (mod p). k= 1 4-2 Primitive roots of composite moduli. Theorem 4-4 imme diately brings the following questions to mind: D o all numbers have primitive roots?
Conversely, if f ( x ) = (x — a)g(x) (mod m), then / ( a ) = 0 (mod m). 42 CONGRUENCES T heorem 3-15 (Lagrange’s theorem). [CHAP. 3 The congruence f(x ) = 0 (mod p) in which f(x ) = a0xn H------ + an, a0 ^ 0 (mod p), has at most n roots. Proof: For n = 1 this follows from Theorem 3-10. Assume that every congruence of degree n — 1 has at most n — 1 solutions, and that a is a root of the original congruence. Then f(x ) = (x - a)q(x ) (mod p), where q(x) is not identically zero (mod p) and is of degree n — 1.
2-5 The least common multiple T heorem 2-7. The number (a, b) = erties: (a) (a, b) > 0; then (a, b)\m. has the following prop- (a, b) (b) a|(a, 6), 6|(a, fe); (c) I f a\m and b|m, 2-5] THE LEAST COMMON MULTIPLE 23 Proof: (b) (a) Obvious. Since (a, 5)|6, we can write (a, b) = \a\ 1H (a, 6) and hence a|(a, b). Similarly, (a,b) = \b\ and so 6|(a, 6). (c) Let lal (a, 6) m = ra = $6, and put d = (a, 6), a = aid, 6 = bid. Then m = raid = sb\d) thus ai|$&i, and since (ai, bi) = 1, it must be that ai|s.