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Math Help - Congruence

  1. #1
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    Congruence

    Prove that a congruent b(modn) if and only if a^2 + b^2 congruent 2ab(mod n^2)

    Help please
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  2. #2
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    Quote Originally Posted by empressA View Post
    Prove that a congruent b(modn) if and only if a^2 + b^2 congruent 2ab(mod n^2)

    Help please


    Hint: a= b\!\!\!\pmod n\Longleftrightarrow a-b= 0\!\!\!\pmod n\Longrightarrow (a-b)^2= 0\!\!\!\pmod {n^2}\Longleftrightarrow ...take it from here (two little steps left).

    Tonio
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  3. #3
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by empressA View Post
    Prove that a congruent b(modn) if and only if a^2 + b^2 congruent 2ab(mod n^2)

    Help please

    a \equiv b(mod n) \Leftrightarrow a^2 + b^2 \equiv 2ab (mod n)

    let's take first \Rightarrow

    we have a \equiv b (mod n) want the right side

     a \equiv b (mod n ) this mean n \mid a-b and this mean there exist an integer x such that a-b = x n
    so

    (a-b)^2 = x^2 n^2

    n^2 \mid (a-b)^2

    (a-b)^2 \equiv 0 (mod n) simplify the left side

    now  \Leftarrow we have

    a^2 + b^2 = 2ab (mod n^2 )

    a^2 - 2ab + b^2 \equiv 0 (mod n^2 )

    (a-b)^2 \equiv 0 (mod n^2 )

    so there exist an integer y such that

    (a-b)^2 = y n^2 note that y is square can you prove it ??

    so we can write y=c^2

    (a-b)^2 = c^2 n^2 by taking square root

    (a-b) = c n

    a-b \equiv 0 (mod n) proof ends
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