# Gaussian integers and norms

1. Feb 22, 2012

### Firepanda

Show that N(a+bi) = even => a+bi divisible by 1+i

So, N(a+bi) = a2+b2 = even

so 2 divides a2+b2

Write 2 = (1+i)(1-i)

so we have 1+i divides a2+b2

so 1+i divides either (a+bi) or (a-bi)

if 1+i divides a+bi we are done

what if 1+i divides a-bi though?

Thats where I'm stuck!

2. Feb 22, 2012

### tiny-tim

hint: what is (1 - i)/(1 + i) ?

(or generally (a + ib)/(1 + i) ?)

3. Feb 22, 2012

### Firepanda

hmm (1 - i)/(1 + i) = -i and (a + ib)/(1 + i) = (1-i)(a+bi)/2

Not sure where you're going with it though!

4. Feb 22, 2012

### tiny-tim

and which of those factors are in Z ?

5. Feb 22, 2012

### Firepanda

so only 1+i is a factor?

6. Feb 22, 2012

### tiny-tim

which of the two factors (1-i)(a+bi)/2 are in Z ?

7. Feb 22, 2012

### Firepanda

Well (1-i) if (a+bi)/2 is the other factor

Or (a+bi) if (1-i)/2 is the other factor

Which should I take?

Last edited: Feb 22, 2012
8. Feb 22, 2012

### tiny-tim

sorry, i've got confused

i meant, is (1-i)(a+bi)/2 in Z ?

9. Feb 22, 2012

### Firepanda

nope!

10. Feb 22, 2012

### tiny-tim

why not?

11. Feb 22, 2012

### Firepanda

Since (a+b)+(bi - ai)/2

So a=b mod 2

i.e can only be a gaussian integer if both a&b are either even or odd?

So it can be a Gaussian integer providing that, and if a&b are even or odd then a2 + b2 = even, which is the norm of a+bi?

12. Feb 22, 2012

### tiny-tim

perfect!

13. Feb 22, 2012

### Dick

I you want just a bit different take on it, then if (1+i) divides (a-bi) take the complex conjugate. So (1-i) divides (a+bi). But (1-i)=(-i)(1+i).