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Math Help - Couldn't solve this one

  1. #1
    Newbie darknight's Avatar
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    Couldn't solve this one

    Positive integers a, b are both relatively prime and less than or equal to 2008. a^2 + b^2 is a perfect square.
    b has the same digits as a in the reverse order. The number of such ordered pairs (a, b) is _________ .

    I started with 2 digits:
    Let a=xy and b=yx

    (10x+y)^2 + (10y+x)^2
    101x^2 + 40xy + 101y^2
    which can't be factorized and isn't a perfect square.

    Tried the same with 3 digits and ended up with this:
    10001x^2 + 200y^2 + 10001z^2 + 400xz + 2020xy + 2020yz
    This also isn't factorizable.

    I've simply got no idea of how to proceed from here.
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  2. #2
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    There simply is NONE. You sure your question is CORRECTLY worded?
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  3. #3
    Newbie darknight's Avatar
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    Quote Originally Posted by Wilmer View Post
    There simply is NONE. You sure your question is CORRECTLY worded?
    Dunno, someone challenged me to solve it. Guess it was his idea of a joke.
    My Apologies..
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  4. #4
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    I am not sure but my answer is zero !

    It is a famous property ( which is not what i am confused ) ,  a-b \equiv 0 \bmod{9} , the proof is as follows :

    Let  a = \sum_{i=0}^n a_i 10^i ~~ a_i \in \{\ 0,1,2,...,9 \}\

    so b =  \sum_{i=0}^n a_i 10^{n-i}  and  a-b = \sum_{i=0}^n  a_i ( 10^i - 10^{n-i} ) \equiv \sum_{i=0}^n a_i ( 1-1) \bmod{9} \equiv 0 \bmod{9}

    We have  a^2 + b^2 = c^2

    Since  a,b coprime , they can be expressed as  m^2 - n^2 ~,~ 2mn , wlog let  a = m^2 - n^2 ~ b = 2mn so we have

     a-b = m^2 - 2mn - n^2 \equiv 0 \bmod{9}

     (m-n)^2 - 2n^2 \equiv 0 \bmod{9}

    I consider the form  x^2 - 2y^2 whether it can be the multiple of 9

    Be caution the quadratic residues are  [R]  = \{\ 0,1,4,7 \} so  2[R] = \{\ 0,2,8,14 \}= \{\ 0,2,5,8 \} , since the intersection of the sets is just   \{\ 0 \} , we conclude  x^2 - 2y^2 \equiv 0 \bmod{9} iff  x \equiv y \equiv 0 \bmod{3}  . Therefore ,  m-n \equiv n \equiv 0 \bmod{3} ~ \implies m \equiv n \equiv 0 \bmod{3}   which is false because ,  (a,b)=1 \implies (m,n)=1 so we can never find out any odered pair (a,b) .


    EDIT: I have made it more complicated , in fact we can consider the quadratic residues from here :

     a^2 + b^2 = c^2


    It is easy to show  a \equiv b \bmod{9} since we have already proved that  a - b \equiv 0 \bmod{9} , so we have :

     2a^2 \equiv c^2 \bmod{9} consider the residues as i mentioned .
    Last edited by simplependulum; June 21st 2010 at 04:05 AM.
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  5. #5
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    Quote Originally Posted by simplependulum View Post
    EDIT: I have made it more complicated , in fact we can consider the quadratic residues from here :
     a^2 + b^2 = c^2
    We could simply apply the "pythagorean triplet" rules, couldn't we, SimpleP ?
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