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Math Help - Sum of two squares

  1. #1
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    Question Sum of two squares

    Do you know how to do this proof?

    Suppose (a , b)=1 . Prove that if p|(a^2+b^2) then p ≡ 1 (mod 4).
    Note: p is any odd prime.
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  2. #2
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    Quote Originally Posted by beta12 View Post
    Do you know how to do this proof?

    Suppose (a , b)=1 . Prove that if p|(a^2+b^2) then p ≡ 1 (mod 4).
    Note: p is any odd prime.
    There are two ways to prove it, the ImPerfectHacker method by examining each case. Or ThePerfectHacker method which is more elegant. Since I am who I say I am I am going to do it my way.

    You have,
    a^2+b^2≡0(mod p)
    Thus,
    a^2≡-b^2(mod P)
    Thus, the Legendre Symbol (I presume you learned them by now),
    Since -b^2 is a quadradic residue of a^2 we have,
    (-b^2/p)=1
    Thus,
    (b^2/p)(-1/p)=1
    But,
    (b^2/p)=1
    Thus,
    (-1/p)=1
    That is a famous case and only happens when,
    p≡1 (mod 4)
    ---
    This looks greatly similar to Fermat's theorem about expressing 4k+1 primes as a sum of two unique squares.
    The proof of that is much more complicated (I wonder how Fermat proved it )
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  3. #3
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    Hi perfecthacker,

    The perfecthacker's way of prove is simple and nice. I like your way very much.

    Do you have the other way to prove this question? Honestly, until now , I still haven't learnt Legendre symbol yet.
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  4. #4
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    Quote Originally Posted by beta12 View Post
    Hi perfecthacker,

    The perfecthacker's way of prove is simple and nice. I like your way very much.

    Do you have the other way to prove this question? Honestly, until now , I still haven't learnt Legendre symbol yet.
    Yes, ImPerfectHacker (my evil twin brother, wait I am the evil twin )
    Would do it like this.

    Divide the prove into cases. (Divison Algorithm)
    'a' can have the form 4k,4k+1,4k+2,4k+3
    Thus,
    'a^2' can only be 4k,4k+1

    'b' can have the form 4k,4k+1,4k+2,4k+3
    Thus,
    'b^2' 4k,4k+1

    The different possible sums between 'a^2+b^2' are,
    4k,4k+1,4k+2
    Since 'p' divides them and is odd it cannot be 4k,4k+2
    Thus, the only possible form of 'a^2+b^2' is 4k+1.

    The prime 'p' can have two forms 4k+1,4k+3
    Now, a number of the form 4k+3 cannot divide a number of the form 4k+1 because they are different forms.
    Thus, p=4k+1
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  5. #5
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    Hi perfecthacker,

    Why do "a" & "b" have the form 4k, 4k+1, 4k+2 , and 4k+3?
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  6. #6
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    Quote Originally Posted by beta12 View Post
    Hi perfecthacker,

    Why do "a" & "b" have the form 4k, 4k+1, 4k+2 , and 4k+3?
    because every number leaves a remainder of 0, 1, 2 or 3 when divided
    by 4. (the k's are not both the same for a and b)

    RonL
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  7. #7
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    Thank you very much.

    I got this question fully.
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