OFFSET
1,1
COMMENTS
For k > 1, a != 1 being a squarefree number (a != -1 unless k is a power of 2), then the ring of integers of Q(a^(1/k)) is Z[a^(1/k)] if and only if: for every p dividing k, we have a^(p-1) !== 1 (mod p^2). In other words, O_Q(a^(1/k)) = Z[a^(1/k)] if and only if none of the prime factors of k is a Wieferich prime of base a. See Theorem 5.3 of the paper of Keith Conrad.
In general, if a^d == 1 (mod p^2) for some d|(p-1), then it is easy to show that x = (1 + a^(d/p) + a^(2*d/p) + ... + a^((p-1)*d/p))/p is an algebraic integer not in Z[a^(1/p)].
Here a = 5. Since 2 is Wieferich prime of base 5, for all even k we have O_Q(a^(1/k)) != Z[a^(1/k)]. There are only 6 other known Wieferich primes of base 5 (A123692) up to 9.7*10^14.
LINKS
Jianing Song, Table of n, a(n) for n = 1..7286 (all terms up to 2*10^8).
Keith Conrad, The ring of integers in a radical extension
EXAMPLE
5^10385 == 1 (mod 20771^2), so x = (1 + 5^(10385/20771) + 5^(2*10385/20771) + ... + 5^(20770*10385/20771))/20771 is an algebraic integer not in Z[5^(1/20771)].
PROG
(PARI) Wieferich_up_to_n(lim) = my(v=[]); forprime(p=2, lim, if(Mod(5, p^2)^(p-1)==1, v=concat(v, p))); v
Up_to_n(lim) = my(pv=Wieferich_up_to_n(lim), list=[]); for(i=2, #pv, my(p=pv[i]); forstep(k=1, lim\p, 2, list=concat(list, k*p))); list=Set(list); list \\ corrected by Jianing Song, Mar 21 2022
CROSSREFS
KEYWORD
nonn
AUTHOR
Jianing Song, Mar 10 2021
STATUS
approved