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

79, 223, 229, 257, 359, 401, 439, 443, 499, 577, 659, 727, 733, 761, 839, 1009, 1087, 1091, 1093, 1129, 1171, 1223, 1229, 1297, 1327, 1367, 1373, 1429, 1489, 1523, 1567, 1601, 1627, 1787, 1811, 1847, 1901, 1907, 1987, 2027, 2029, 2081, 2089, 2099, 2143, 2153, 2207, 2213, 2251, 2399, 2459, 2467

Offset: 1

Author

Alonso del Arte, Nov 28 2016

Keywords

• nonn

Comments

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).

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.

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.

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.

Such an obvious example of multiple distinct factorizations is obviously not available in O_(Q(sqrt(p))).

Examples

			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.
Z[sqrt(83)] is a unique factorization domain, hence 83 is not in the sequence.
		

Links

• G. C. Greubel, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A146209.

Programs

Mathematica

Select[Prime[Range[100]], NumberFieldClassNumber[Sqrt[#]] > 1 &]

Extensions

Missing term 2089 added by Emmanuel Vantieghem, Mar 08 2019