OFFSET
1,1
COMMENTS
Using Fermat's little theorem twice, it is easy to see that m=p^2-1 solves this problem for all odd primes p. In fact, there appear to be exactly p-1 values of m with 1 <= m <= p^2 for which m^(p-1) == 1 (mod p^2). See A096082 for the related open problem. - T. D. Noe, Aug 24 2008
That there are exactly p-1 values of 1 <= m <= p^2 for which m^(p-1) == 1 (mod p^2) follows immediately from Hensel's lifting lemma and Fermat's little theorem - every solution mod p corresponds to a unique solution mod p^2. - Phil Carmody, Jan 10 2011
For n > 2, prime(n) does not divide a(n)^2 - 1, so a(n) is the smallest m > 1 such that (m^(prime(n)-1) - 1)/(m^2 - 1) == 0 (mod prime(n)^2). - Thomas Ordowski, Nov 24 2018
REFERENCES
P. Ribenboim, The New Book of Prime Number Records, Springer, 1996, 345-349.
LINKS
T. D. Noe, Table of n, a(n) for n = 1..10000
EXAMPLE
For n=3, p=5 is the third prime and 5^2 = 25 divides 7^4 - 1 = 2400.
MATHEMATICA
dpa[n_]:=Module[{p=Prime[n], a=2}, While[PowerMod[a, p-1, p^2]!=1, a++]; a]; Array[dpa, 70] (* Harvey P. Dale, Sep 05 2012 *)
PROG
(PARI) a(n) = my(p=prime(n)); for(a=2, oo, if(Mod(a, p^2)^(p-1)==1, return(a))) \\ Felix Fröhlich, Nov 24 2018
(Python)
from sympy import prime
from sympy.ntheory.residue_ntheory import nthroot_mod
def A039678(n): return 2**2+1 if n == 1 else int(nthroot_mod(1, (p:= prime(n))-1, p**2, True)[1]) # Chai Wah Wu, May 18 2022
CROSSREFS
KEYWORD
nonn,nice
AUTHOR
EXTENSIONS
More terms from David W. Wilson
Definition adjusted by Felix Fröhlich, Jun 24 2014
Edited by Felix Fröhlich, Nov 24 2018
STATUS
approved