 By Leonard D. Baumert

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Additional info for Cyclic Difference Sets

Example text

30) it follows that it is not only K(~q). 32) s 8(~ ) = ~q e(~q). 32) that t t ~q -~ ~q since, as noted above, contradiction, so t nl~q. 33) ~ nl~q~. ) W~q* into - ~q*. 33) is a necessarily is a non-zero square modulo q. 9 only the last one remains to be verified. If the prime ql ~ q were another (necessarily odd) prime divisor of which there existed a multiplier modulo ql' tI and integer fl such that then, by the process used above, it would follow that 23 n0q (since it cannot be both and 23 m0ql).

Mod x v - l) is a multiplier of this difference set as was to be proved. E. above. generalization. known case. 1 represents a further (n0~v) = 1 are superfluous in every For~ no cyclic difference sets are known with prime divisor p of n (n,v) > 1 and every is a multiplier of every known cyclic difference set. Morris Newman (1963) extended this result slightly by showing that the odd prime p is always a multiplier whenever n = 2p and (7p,v) = 1. Turyn (1964) generalized Newman's result. 1A (Turyu) that for every prime p Let dividing n = 2n 0 nO t nO odd and prime to there is an integer pJP -= t Then with modulo jp v.

Prime divisors of where ~ (Yamamoto, 1963) v and let q~ Let rm denotes Euler's function. q = 4t + 3 and r v with strictly divide be distinct odd (\$(q~)~ ~(rm)) = 2, Assume that any prime divisor p of n modulo r satisfies one of the conditions: (i) Ordqp ~ 0 (mod 2) (ii) order of p and modulo q~ OrdrP ~ 0 (mod 2) ~ 0(mod 4), is ½ ~(q~) and order of p m is ~(rm), (iii) p = q and order of p Then, if there exists a non-trivial 4n = x 2 + qy 2 , 0 ~ x, has a solution in integers modulo rm is \$(rm). v,k,h - difference set, the equation 0 ~ y ~ 2vq-~r-m, x + y 4vq-~r-m x, y.