# Quadratic integers help

1. Jun 16, 2010

### squelchy451

For the theorem that states that in quadratic field Q[sqrt d], if d is congruent to 1 mod 4, then it is in the form (a + b sqrt d)/2 and if it's not, it's in the form a + b sqrt d where a and b are rational integers, is it saying that if a and b are rational integers and the quadratic number are in the form according to its congruency mod 4, then the quadratic number is an integer?

Also, how would you prove that if 32 = ab where a and b are relatively prime quadratic integers in Q[sqrt -1], a = e(g^2) where e is a unit and g is a quadratic integer in Q [sqrt -1].

2. Jun 18, 2010

### ramsey2879

When you have the product of two quadratic integers equal an integer, look first for two of the form A +/- B(sqrt -1). That is one use negative B and the other uses positive B so that the quadratic part cancels out.

Last edited: Jun 18, 2010
3. Jun 20, 2010

### squelchy451

On a side note, are a + b sqrt d and a - b sqrt d where a, b, and d are rational integers (and d is not a perfect square) relatively prime?

4. Jun 20, 2010

### ramsey2879

Well if 2, b and sqrt d each are relatively prime to a, then I would say that the two factors are relatively prime but I am not sure. However, I think you have to look to cancel the quadratic part in another way since 4 + 4i and 4 - 4i are not relatively prime. Maybe try playing around with numbers 1,16 and i in lieu of 4 and i.

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