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A256567
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Primes p with the property that there are three consecutive integers (x,x+1,x+2) with 1 < x <= p-3 whose product is 1 modulo p.
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6
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7, 11, 17, 19, 23, 37, 43, 53, 59, 61, 67, 79, 83, 89, 97, 101, 103, 107, 109, 113, 137, 149, 157, 167, 173, 181, 191, 199, 211, 223, 227, 229, 241, 251, 263, 271, 281, 283, 293, 307, 313, 317, 337, 347, 359, 367, 373, 379, 383, 389, 401, 419, 421, 431, 433, 449
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OFFSET
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1,1
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COMMENTS
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There may be one or more such triples, but 23 is the only prime up to 100000 having precisely two such triples. For the number of triples for each prime, see A256572.
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LINKS
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EXAMPLE
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For p=7: 4*5*6=120==1 (mod 7), so 7 is a term.
For p=11: 5*6*7=210==1 (mod 11), so 11 is a term.
For p=17: 4*5*6=120==1 (mod 17), so 17 is a term.
13 is not a term because there is no such triple with product == 1 (mod 13).
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PROG
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(R)
library(numbers)
IP <- vector()
t <- vector()
S <- vector()
IP <- c(Primes(1000)) # Build a vector of all primes < 1000.
for (j in 1:(length(IP))){
for (i in 3:(IP[j]-2))
t[i-1] <- as.vector(mod(((i-1)*i*(i+1)), IP[j]))
S[j] <- length(which(t==1))
}
IP[S!=0]
#The loop checks for every triple for every prime, what it is modulo that prime. "IP[S!=0]" lists the primes that have at least one triple. For all p<10000 it takes a few minutes. For all p<100000 a few hours.
(PARI) isok(p) = {if (isprime(p), for (x=1, p-3, if (Mod(x*(x+1)*(x+2), p) == 1, return (1)); ); ); } \\ Michel Marcus, Oct 05 2021
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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