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Re: Sum of four squares

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  • aldimalkoun
    ... the ... Well, since there hasn t been many replies to this problem, I should perhaps say some more about it. The first remark is this: for the sum of two
    Message 1 of 3 , Sep 2, 2002
      --- In mathforfun@y..., "aldimalkoun" <joemalkoun@h...> wrote:
      > Now here goes the famous problem: every integer can be written as
      the
      > sum of (at most) four squares. Can you prove it?

      Well, since there hasn't been many replies to this problem, I should
      perhaps say some more about it. The first remark is this: for the sum
      of two squares problem, it has been useful to work in the ring of
      integral complex numbers, or Gaussian integers. For the sum of 4
      squares, it's probably useful to work in the ring of integral
      quaternions.

      For the sum of two squares problem, one can use the fact that the
      product of two numbers, each of them being the sum of two squares, is
      itself the sum of two squares.

      For the sum of four squares problem, there is a similar fact: the
      product of two numbers, each of them being the sum of four squares, is
      itself the sum of four squares.

      Both of these facts derive from the formula: N(ab)=N(a)*N(b), where
      a,b are integral complex numbers or integral quaternions, and where
      N(a) is the square of the modulus of a. Example: N(2+5i)=2^2+5^2=29.

      Thus, since 2 can be written as the sum of four squares
      (2=1^2+1^2+0^2+0^2), and a prime of the form 4k+1 can be written as
      the sum of two squares, and hence as the sum of four squares, it
      suffices to show that any prime of the form 4k+3 can be written as the
      sum of four squares.
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