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Math Help - Prime 2

  1. #1
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    Prime 2

    Hi,

    How to show that if ab=cd \Longrightarrow      a^2+b^2+c^2+d^2 is not prime????
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  2. #2
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    Are you sure this is true?
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  3. #3
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    Quote Originally Posted by elim View Post
    Are you sure this is true?
    Do you know a counterexample? I can't think of one myself.
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  4. #4
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    Quote Originally Posted by bhitroofen01 View Post
    Hi,

    How to show that if ab=cd \Longrightarrow a^2+b^2+c^2+d^2 is not prime????
    let \gcd(a,c)=x, \ \gcd(b,d)=y. then a=xa_0, \ c=xc_0, \ b=yb_0, \ d=yd_0, where \gcd(a_0,c_0)=\gcd(b_0,d_0)=1. thus, since ab=cd, we must have a_0=d_0 and b_0=c_0.

    therefore a^2+b^2+c^2+d^2=(x^2+y^2)(a_0^2+b_0^2). \ \Box
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  5. #5
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    Let p=(a,c), then a=pr,\;c=ps,\;(r,s) = 1

    Since ab=cd,\; prb=psd, so rb=sd. \; s|b

    Assume b=qs, from rb=sd we get rqs=sd or d=qr

    Now put a=pr,\; b=qs,\; c=ps,\; d=qr into a^2+b^2+c^2+d^2 we get

    a^2+b^2+c^2+d^2=(pr)^2+(qs)^2+(ps)^2+(qr)^2

    =p^2 r^2+p^2 s^2+q^2 r^2+q^2 s^2 =(p^2+q^2)(s^2+r^2)

    Where p^2+q^2, s^2+r^2 >1 and so a^2+b^2+c^2+d^2 is not prime
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  6. #6
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    I am curious if this statement is correct ...

    If a prime can be expressed as the sum of two squares  x^2 + y^2 , then ( |x| , |y|) is the unique pair .

    If it is true , we can solve it in this way :

     A =  a^2 + b^2 + c^2 + d^2 = (a-b)^2 + (c+d)^2 = (a+b)^2 + (c-d)^2

    If  a+b = c+d , then  a^2 + b^2 = c^2 + d^2 so  A is an even number ( not 2 , not prime ) .

    Therefore , A has at least two pairs and it is not prime .
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