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Math Help - Sums of squares exercise

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    Sums of squares exercise

    Let  m_0 be the least possible  m for which there are x and y for which

    x^2 + y^2 = mp , where p is a prime and p \equiv 1 (mod4) . Also p \nmid x , p \nmid y ,  0<m<p .

    That there are such x,y and m is the subject of the previous exercise.

    Thus, x_0^2+y_0^2 = m_0p (1)

    Show m_0 \nmid x_0 and  m_0 \nmid y_0 unless  m_0 = 1 .

    By taking  m_0 as a factor of both  x_0 and  y_0 , I find, from (1) above, that  p must be factored by  m_0 . But p is prime. Therefore m_0=1 .

    I feel I must have overlooked something, because I have not used the hypothesis that  m_0 is minimal. What have I overlooked, please?

    The above problem is from GE Andrews's "Number Theory", ex. 2, Chapter 11-2.
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  2. #2
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    Quote Originally Posted by Nivla View Post
    Let  m_0 be the least possible  m for which there are x and y for which

    x^2 + y^2 = mp , where p is a prime and p \equiv 1 (mod4) . Also p \nmid x , p \nmid y ,  0<m<p .

    That there are such x,y and m is the subject of the previous exercise.

    Thus, x_0^2+y_0^2 = m_0p (1)

    Show m_0 \nmid x_0 and  m_0 \nmid y_0 unless  m_0 = 1 .



    By taking  m_0 as a factor of both  x_0 and  y_0 , I find, from (1) above, that  p must be factored by  m_0 . But p is prime. Therefore m_0=1 .

    I feel I must have overlooked something, because I have not used the hypothesis that  m_0 is minimal. What have I overlooked, please?

    The above problem is from GE Andrews's "Number Theory", ex. 2, Chapter 11-2.
    without loss of generality suppose that m_0 \mid x_0. if m_0 > 1, then \gcd(m_0,y_0)=d > 1 because m_0 \mid m_0p - x_0^2=y_0^2. now let x_0=m_0x_1, \ m_0=dm_1, \ y_0=dy_1.

    then x_0^2+y_0^2=m_0p will give us d(m_1^2x_1^2 + y_1^2)=m_1p. thus d \mid m_1. let m_1=dm_2. then (m_1x_1)^2 + y_1^2=m_2p, which contradicts minimality of m_0 because m_0=d^2m_2 > m_2.
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    I really appreciate your insight and made a small donation for your help. Thanks, again!
    Nivla
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