Posted by: "djbroadhurst" d.broadhurst@... djbroadhurst

Date: Sat Oct 3, 2009 3:53 pm ((PDT))

All this is very simply explained

by unique factorization in K(sqrt(-1)).

1) A number N is the sum of two squares if and only if

every prime divisor of N that is congruent to 3 mod 4

occurs to an even power.

2) An odd number N is the sum of two coprime squares in precisely

one way if only if N is a prime congruent to 1 mod 4

or a power of such a prime.

************

David, a Minor correction is needed here.

Note that cubes of primes congruent to 1 mod 4 do not have unique sum of square decomposition.

(2 + i)(2+i)(2+ i) = (3 + 4 i) (2+i) = 2 + 11 i

2**2 + 11**2 = 5**3

(2 + i)(2+i)(2-i) = (2 + i) * 5 = 10 + 5 i

10**2 + 5**2 = 5**3

************

Note that (1) is a consequence of Theorem 252 of Hardy and Wright,

that the primes of K(sqrt(-1)) are: 1 + I, the rational primes

congruent to 3 mod 4, the factors a + b*I of the rational primes

congruent to 1 mod 4, and the associates of these primes.

To understand (2), suppose that

p = norm(a+I*b) = a2 + b2 and

q = norm(c+I*d) = c2 + d2

are distinct rational primes congruent to 1 mod 4.

Then we may write

p*q = norm((a+I*b)*(c+I*d)) = (a*c-b*d)2 + (b*c+a*d)2

p*q = norm((a+I*b)*(c-I*d)) = (a*c+b*d)2 + (b*c-a*d)2

in two essentially different ways.

But for any power k of p, the only sum of coprime squares is

p^k = norm((a+I*b)^k)

since if we try something like

p^k = norm((a+I*b)^(k-j)*(a-I*b)^j)

with k > j > 0, then p will divide both of the resultant squares.

David