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unique sum square decomposition

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  • Kermit Rose
    1g. Re: Formula Posted by: djbroadhurst d.broadhurst@open.ac.uk djbroadhurst Date: Sat Oct 3, 2009 3:53 pm ((PDT)) All this is very simply explained by
    Message 1 of 1 , Oct 4, 2009
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      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.

      David
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