unique sum square decomposition
- 1g. Re: Formula
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.