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Math Help - Prove this proposition

  1. #1
    Super Member dhiab's Avatar
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    Prove this proposition

    Prove :
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  2. #2
    Senior Member TriKri's Avatar
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    here's a solution:

    Suppose \frac{a^2+b^2}{a^2-b^2} is an integer. Then we have

    \frac{a^2+b^2}{a^2-b^2} = \frac{a^2-b^2}{a^2-b^2}+\frac{2b^2}{a^2-b^2} = 1+\frac{2b^2}{a^2-b^2} \in \mathbb{N}\Leftrightarrow\frac{2b^2}{a^2-b^2} \in \mathbb{N}

    \Leftrightarrow 2b^2 = n(a^2-b^2), n\in\mathbb{N} \Leftrightarrow (n+2)b^2 = na^2 \Leftrightarrow

    Set p=\frac{a}{gcd(a,\ b)} and q=\frac{b}{gcd(a,\ b)}, where gcd(a, b) is the greatest common divisor of a and b. Note that both p and q are integers but that they don't have any common prime factor.

    (n+2)b^2 = na^2\Leftrightarrow(n+2)q^2=np^2\Leftrightarrow\fr  ac{n+2}{p^2}=\frac{n}{q^2}=\frac{j}{k},\ j,\ k \in\mathbb{N},\ gcd(j,\ k) = 1

    Every prime factor in k also has to be in p^2 and q^2, but since gcd(p,\ q)=1, p^2 and q^2 share no prime factors. So k = 1

    \therefore\ n+2=jp^2\text{ and }n=jq^2

    p^2=q^2+\frac{2}{j} < q^2+2q+1=(q+1)^2

    a > b\Leftrightarrow p > q

    \therefore\ q < p < q+1

    But there are no integers in the interval (q, q+1).

    So \frac{a^2+b^2}{a^2-b^2} cannot be an integer.
    Last edited by mr fantastic; October 22nd 2009 at 06:06 AM.
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  3. #3
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    Simpler

    Suppose \frac{a^2+b^2}{a^2-b^2}=n\in\mathbb{N}. (Start by noticing n\neq1 else b=0.) Rearranging, \frac{a^2}{b^2}=\frac{n+1}{n-1}=\frac{n^2-1}{(n-1)^2}. So n^2-1=\frac{a^2}{b^2}(n-1)^2. Since n\neq1, the RHS is a perfect nonzero square. The LHS is one less than a perfect square. No two squares differ by only one in the naturals. QED
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  4. #4
    Senior Member TriKri's Avatar
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    Nice.
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