In the field of real numbers the exponential function exp(x) is the sum of the quantities (x^k)/k! for k ranging over all the distinct non-negative integers. In the finite field of integers modulo a prime p we can define the analagous function where k ranges over all the distinct non-negative residues. In other words, given any prime p, and letting LNNR[x] denote the least non-negative residue of x modulo p, we define the function _ _ | x^2 x^3 x^(p-1) | a(x) = LNNR| 1 + x + --- + ---... + ------- | |_ 2! 3! (p-1)! _| This function involves divisions (mod p), so it would be tricky to evaluate if the modulus p was composite. However, the companion function b(x) = LNNR[ 1 + (1!)x + (2!)x^2 + (3!)x^3 + ... ((p-1)!)x^(p-1) ] can be easily computed for any modulus p, prime or composite. Also, it turns out that a(x) = -b(-1/x) (mod p) for any prime p and any x not congruent to zero (mod p). Thus, for any integer modulus we can simply define a(x) to be LNNR[-b(-1/x)] for all non-zero x, and of course a(0)=1. If, for a given prime modulus p, we let A(p) denote the sum of the values of a(x) for x = 0 to p-1, and similarly let B(p) denote the sum of all the values of b(x), then we find that A(p) = B(p) = 1 (mod p) We can also consider the values of p for which the absolute sums A(p) and B(p) are equal. The only primes less than 1000 with this property are 2, 3, 17, 23, 43, 71, 109, 337, 769, 853, 919 Anyway, for a given modulus m let z(m) denote the number of "zeros" of a(x). In other words, z(m) is the number of distinct values of x for which a(x)=0. I think z(m) is multiplicative, i.e., if gcd(r,s)=1 then z(rs)=z(r)z(s). For this reason we need only consider the values of z(p) for primes p. It appears that z(p^k) = z(p) for all exponents k. The values of z(p) for the first several primes are p z(p) p z(p) p z(p) p z(p) 2 1 53 3 127 0 199 3 3 0 59 1 131 1 211 0 5 1 61 0 137 0 223 2 7 1 67 1 139 1 227 1 11 1 71 2 149 0 229 3 13 2 73 1 151 5 233 3 17 0 79 1 157 0 239 2 19 1 83 1 163 2 241 2 23 0 89 2 167 4 251 0 29 0 97 4 173 1 257 0 31 1 101 2 179 0 263 1 37 4 103 1 181 1 269 0 41 1 107 0 191 3 271 0 43 2 109 0 193 2 277 0 47 2 113 0 197 1 281 0 Notice that z(151)=5. This is the only prime p less than 1000 such that z(p) is greater than 4. For the 168 primes less than 1000 the values of z(p) are distributed as follows z(p) 0 1 2 3 4 5 --- --- --- --- --- --- # of primes 54 60 40 9 4 1 Does anyone know the asymptotic distribution over all primes? Is there a prime with 6 zeros? Letting z(p) denote the number of distinct solutions of x^2 x^3 x^(p-1) 1 + x + --- + ---... + ------- = 0 (mod p) 2! 3! (p-1)! does there exist a prime p such that z(p)=n for any given n? I've checked up to 6863 and the only primes in this range (other than 151) with more than four roots are 5209 and 5653, each of which has five roots. (I note in passing that 5209 = -1/2 (mod 151), and 5653 = -1/16 (mod 151).) There is no prime in this range with more than five roots. As for the asymptotic distribution, it appears (not surprisingly) to be Poisson with expected value of 1. The values of z(p) for the 883 primes up to 6863 are distributed as follows 0 1 2 3 4 5 6 --- --- --- --- --- --- --- Actual 283 355 183 48 12 3 0 Poisson 324 324 162 54 13 2.7 .451 This seems to suggest that there is probably a 6-root prime less than, say, 20000. As to whether there is a prime p with z(p)=n for every integer n, that seems less clear. If the distribution really is Poisson with expected value 1, then the density of primes with z(p)=n is 1/(e*n!), which would make primes with, say, 100 roots extrodinarily rare. It would be interesting to know how far the Poisson model accurately predicts the density of z(n). It would also be interesting to know if there is an efficient way of computing z(p) for a given prime p (i.e., more efficient than just testing all the residue classes one by one). Also, is there a way of "constructing" primes with a large number of zeros? UPDATE: On 17 June 2000 Don Reble sent an email reporting that he had found the smallest prime p such that z(p) equals 6. The prime is p = 11117, for which the six roots of a(x) are 1500 3380 3897 4156 6694 9949 He checked all primes less than 32768, and this is the only one with z(p) greater than 5 in this range. He also identified several more primes with z(p)=5, so the complete list of all known such primes is now 151 5209 5653 7639 9343 9871 11329 12373 12979 14731 15461 16267 17159 19457 26591 27281 28309 32467 Over the range of all primes less than 32768, Don found that the distribution of primes according to their values of z(p) is as shown below # of primes predicted less than based on z(p) 32768 Poisson ---- -------- -------- 0 1249 1292 1 1317 1292 2 643 646 3 228 215 4 56 54 5 18 10.76 6 1 1.79 7 0 0.256 8 0 0.032 This certainly seems consistent with the premise that the distribution is Poisson, i.e., the fraction of primes with z(p)=n is asymptotic to 1/(e*n!).

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