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Thread: Sums of squares exercise

  1. #1
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    Sums of squares exercise

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

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

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

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

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

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

    I feel I must have overlooked something, because I have not used the hypothesis that $\displaystyle 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 $\displaystyle m_0 $ be the least possible $\displaystyle m $ for which there are $\displaystyle x$ and $\displaystyle y$ for which

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

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

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

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



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

    I feel I must have overlooked something, because I have not used the hypothesis that $\displaystyle 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 $\displaystyle m_0 \mid x_0.$ if $\displaystyle m_0 > 1,$ then $\displaystyle \gcd(m_0,y_0)=d > 1$ because $\displaystyle m_0 \mid m_0p - x_0^2=y_0^2.$ now let $\displaystyle x_0=m_0x_1, \ m_0=dm_1, \ y_0=dy_1.$

    then $\displaystyle x_0^2+y_0^2=m_0p$ will give us $\displaystyle d(m_1^2x_1^2 + y_1^2)=m_1p.$ thus $\displaystyle d \mid m_1.$ let $\displaystyle m_1=dm_2.$ then $\displaystyle (m_1x_1)^2 + y_1^2=m_2p,$ which contradicts minimality of $\displaystyle m_0$ because $\displaystyle m_0=d^2m_2 > m_2.$
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  3. #3
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    I really appreciate your insight and made a small donation for your help. Thanks, again!
    Nivla
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