By Euler L.

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**Extra info for A commentary on the continued fraction by which the illustrious La Grange has expressed the binomial powers**

**Sample text**

The ring of integers of K = Q( −6) is OK = Z[ −6]. In OK , we have √ √ 6 = − −6 −6 = 2 · 3. √ If ab = −6, with a, b ∈ OK and neither a unit, then Norm(a) Norm(b) = 6, so √ without loss Norm(a) = 2 and Norm(b) = 3. If a = c + d −6, then Norm(a) = c2 + 6d2 ; since the equation c2 + 6d2√= 2 has no solution with√c, d ∈ Z, there is no element in OK with norm 2, so −6 is irreducible. Also, −6 is not a unit times 2 or times 3, since again the norms would not match up. Thus 6 can not be written uniquely as a product of irreducibles in OK .

We will use Magma, which implements the algorithm described in the previous section, to show that 2 is an essential discriminant divisor for K. 56 CHAPTER 8. FACTORING PRIMES > K := NumberField(x^3 + x^2 - 2*x + 8); > OK := MaximalOrder(K); > Factorization(2*OK); [

11. The ring of integers of K = Q( −6) is OK = Z[ −6]. In OK , we have √ √ 6 = − −6 −6 = 2 · 3. √ If ab = −6, with a, b ∈ OK and neither a unit, then Norm(a) Norm(b) = 6, so √ without loss Norm(a) = 2 and Norm(b) = 3. If a = c + d −6, then Norm(a) = c2 + 6d2 ; since the equation c2 + 6d2√= 2 has no solution with√c, d ∈ Z, there is no element in OK with norm 2, so −6 is irreducible. Also, −6 is not a unit times 2 or times 3, since again the norms would not match up. Thus 6 can not be written uniquely as a product of irreducibles in OK .