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Math Help - Problem 35

  1. #1
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    Problem 35

    The mathematician, Fermat, said that any prime p of the form 4k+1 can be expressed as a sum of two squares. However, it turns out amazingly, that this representation is also unique. Accept by faith the existence of Fermat's theorem and prove uniqueness.
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    I think I have got a solution, but's its also prooving the existence of the two numbers : no acceptation by faith!
    would it fit?
    (i hope not, because i feel lazy and though it's not very long it's a little tricky because it's involving the distinction betwenn numbers and their 'correspondant' modulo 4k+1=p')
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    Say that p = a^2+b^2 = c^2 + d^2.

    Note that a^2d^2 - b^2c^2 = (p-b^2)d^2 - b^2(p-d^2) = p(d^2-b^2)\equiv 0 (\bmod p).
    Thus,
    (ad - bc)(ad+bc)\equiv 0 (\bmod p) \mbox{ thus }ad -bc = 0 \mbox{ or }ad+bc = p.

    Our goal now is to show either ad=bc \mbox{ or }ac=bd. Now ad=bc follows at once from the equation ad -bc =0. But if ad+bc=p we will show this implies that ac=bd.

    Now if p=ad+bc then p^2 = (a^2+b^2)(c^2+d^2) = (ad+bc)^2+(ac-bd)^2 = p^2+(ac-bd)^2. Thus, ac=bd.

    Now we have established that ad=bc \mbox{ or }ac=bd.

    Say the first one is true ad=bc. Then a|bc. But since \gcd(a,b)=1\implies a|c. Say therefore that c=ak. Substitute c=ak into ad=bc we get ad = abk \implies d=bk. But then p = c^2+d^2 = (ak)^2+(bk)^2 = k^2(a^2+b^2)\implies k=1 \implies a=c \mbox{ and }b=d.

    Now using a similar argument with ac=bd we find that a=d \mbox{ and }b=c.

    I hope you like this proof. I really find it beautiful how divisibility arguments lead us to this conclusion.
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  4. #4
    MHF Contributor Bruno J.'s Avatar
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    Suppose p=a^2+b^2=c^2+d^2.
    Then
    p=(a+bi)(a-bi)=(c+di)(c-di)

    and each of a \pm bi, c \pm di are primes in \mathbb{Z}[i] because they have prime norms.
    But \mathbb{Z}[i] is a unique factorization domain hence the primes are the same up to unit factors and rearrangement (i.e. : a+bi \in \{\pm(c+di), \pm i (c+di)\}. This yields that \{a^2,b^2\}=\{c^2,d^2\} and we are done.
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