OFFSET
1,5
COMMENTS
Conjecture: a(n) > 0 for all n > 3. In other words, for any prime p > 5, there exists a positive square k^2 < p such that the inverse of k^2 mod p among 1, ..., p-1 is prime.
We have verified this for all n = 4, ..., 10^7. See also A242441 for an extension of the conjecture.
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
EXAMPLE
a(4) = 1 since 2^2*2 == 1 (mod prime(4)=7).
a(6) = 1 since 3^2*3 == 1 (mod prime(6)=13).
a(9) = 1 since 4^2*13 == 1 (mod prime(9)=23).
a(18) = 1 since 7^2*5 == 1 (mod prime(18)=61).
a(30) = 1 since 8^2*83 == 1 (mod prime(30)=113).
a(46) = 1 since 10^2*2 == 1 (mod prime(46)=199).
a(52) = 1 since 15^2*17 == 1 (mod prime(52)=239).
a(97) = 1 since 18^2*11 == 1 (mod prime(97)=509).
MATHEMATICA
r[k_, n_]:=PowerMod[k^2, -1, Prime[n]]
Do[m=0; Do[If[PrimeQ[r[k, n]], m=m+1], {k, 1, Sqrt[Prime[n]-1]}]; Print[n, " ", m]; Continue, {n, 1, 80}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, May 13 2014
STATUS
approved