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Thread: Proof validation

  1. #1
    Junior Member
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    Dec 2005
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    Proof validation

    I need to prove that if $\displaystyle a,b$ are integers then $\displaystyle a^2-b^2=101010$ is not possible.

    My proof is this:
    Since 101010 can be divided with 5 which is prime number, then one of factors $\displaystyle (a-b)(a+b)$ also must be 5.
    Lets assume that $\displaystyle a-b=5$ and $\displaystyle a+b=5k$.
    Because it must be $\displaystyle (a-b)(a+b)=101010$ then it must be $\displaystyle a+b=20202$.
    So, we have to solve system of equations in order to see whether $\displaystyle a,b$ are integers.

    $\displaystyle \begin{array}{l}
    \left\{ \begin{array}{l}
    a - b = 5 \\
    a + b = 20202 \\
    \end{array} \right. \\
    \\
    \left\{ \begin{array}{l}
    a = 5 + b \\
    a = 20202 - b \\
    \end{array} \right. \\
    \\
    5 + b = 20202 - b \\
    2b = 20202 - 5 \\
    b = \frac{{20197}}{2} \\
    \end{array}
    $

    Since $\displaystyle \frac{{20197}}{2}$ is not integer then b is not integer.

    Is this proof valid?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by DenMac21
    I need to prove that if $\displaystyle a,b$ are integers then $\displaystyle a^2-b^2=101010$ is not possible.

    My proof is this:
    Since 101010 can be divided with 5 which is prime number, then one of factors $\displaystyle (a-b)(a+b)$ also must be 5.
    Lets assume that $\displaystyle a-b=5$ and $\displaystyle a+b=5k$.
    101010 divisible by 5 implies one of (a-b) and (a+b) is divisible by 5, not
    that one of them is 5. Also if one were 5 the other need not be divisible
    by 5.

    Because it must be $\displaystyle (a-b)(a+b)=101010$ then it must be $\displaystyle a+b=20202$.
    So, we have to solve system of equations in order to see whether $\displaystyle a,b$ are integers.

    $\displaystyle \begin{array}{l}
    \left\{ \begin{array}{l}
    a - b = 5 \\
    a + b = 20202 \\
    \end{array} \right. \\
    \\
    \left\{ \begin{array}{l}
    a = 5 + b \\
    a = 20202 - b \\
    \end{array} \right. \\
    \\
    5 + b = 20202 - b \\
    2b = 20202 - 5 \\
    b = \frac{{20197}}{2} \\
    \end{array}
    $

    Since $\displaystyle \frac{{20197}}{2}$ is not integer then b is not integer.

    Is this proof valid?
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