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A278837 Primes p such that the ring of algebraic integers of Q(sqrt(p)) does not have unique factorization.

%I A278837 #28 Mar 08 2025 04:05:22
%S A278837 79,223,229,257,359,401,439,443,499,577,659,727,733,761,839,1009,1087,
%T A278837 1091,1093,1129,1171,1223,1229,1297,1327,1367,1373,1429,1489,1523,
%U A278837 1567,1601,1627,1787,1811,1847,1901,1907,1987,2027,2029,2081,2089,2099,2143,2153,2207,2213,2251,2399,2459,2467
%N A278837 Primes p such that the ring of algebraic integers of Q(sqrt(p)) does not have unique factorization.
%C A278837 It is still unknown whether there are infinitely many real, positive, squarefree d such that O_(Q(sqrt(d))) has unique factorization (or, to put it another way, the class number is 1).
%C A278837 If one only looks at small prime numbers, one could easily be tempted to think that if p is prime then O_(Q(sqrt(p))) has unique factorization.
%C A278837 By contrast, given distinct primes p and q, one could think that O_(Q(sqrt(p*q))) generally does not have unique factorization, especially if p = 5.
%C A278837 It then often happens that both p and q are irreducible, and therefore p*q = (sqrt(p*q))^2 represents two distinct factorizations of the same number.
%C A278837 Such an obvious example of multiple distinct factorizations is obviously not available in O_(Q(sqrt(p))).
%H A278837 G. C. Greubel, <a href="/A278837/b278837.txt">Table of n, a(n) for n = 1..1000</a>
%e A278837 In Z[sqrt(79)], to pick just one example of a number having more than one distinct factorization, we verify that 3 and 5 are both irreducible, yet 15 = 3 * 5 = (-1)*(8 - sqrt(79))*(8 + sqrt(79)). Thus 79 is in the sequence.
%e A278837 Z[sqrt(83)] is a unique factorization domain, hence 83 is not in the sequence.
%t A278837 Select[Prime[Range[100]], NumberFieldClassNumber[Sqrt[#]] > 1 &]
%Y A278837 Cf. A146209.
%K A278837 nonn
%O A278837 1,1
%A A278837 _Alonso del Arte_, Nov 28 2016
%E A278837 Missing term 2089 added by _Emmanuel Vantieghem_, Mar 08 2019