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    Let F be a field (with 2\neq 0) and define:

    D_F(2)=\{a\in F^* : a \ \text{is a sum of} \ 2 \ \text{squares in} \ F\}

    I'm trying to prove that D_F(2) is a group with multiplication. This identity shows closure:

    (x_1^2+x_2^2)\cdot(y_1^2+y_2^2)=  (x_1 y_1 - x_2 y_2)^2+ (x_2 y_1+x_1 y_2)^2

    Can someone help with finding the inverse?
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    Hold on, you have two binary operations (addition, multiplication) on F? Is F meant to be a field (or a ring)? "Sum of squares" doesn't really make much sense in the context of a group.
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    (Original post by nuodai)
    Hold on, you have two binary operations (addition, multiplication) on F? Is F meant to be a field (or a ring)? "Sum of squares" doesn't really make much sense in the context of a group.
    Typo sorry. (Edited).
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    (Original post by 0-))
    Typo sorry. (Edited).
    Ah alright. Well you can write \dfrac{1}{x_1^2+x_2^2} as \dfrac{x_1^2+x_2^2}{(x_1^2+x_2^2  )^2}. How else can this be written?
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    (Original post by nuodai)
    Ah alright. Well you can write \dfrac{1}{x_1^2+x_2^2} as \dfrac{x_1^2+x_2^2}{(x_1^2+x_2^2  )^2}. How else can this be written?
    As the sum of two squares!
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    (Original post by 0-))
    As the sum of two squares!
    Correct :p:
 
 
 
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