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Math Help - lovely result, but curious

  1. #1
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    Omagh
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    lovely result, but curious

    Hi,

    Whilst meandering through integers that fit the equation

    a^2+b^2=2c^2

    I came across this curious result

    1^2+7^2=2(5^2)
     11^2+77^2=2(55^2)
     111^2+777^2=2(555^2)

    I went as far as 16 digits and it still holds true.

     1111111111111111^2+7777777777777777^2=2(5555555555  555555^2)

    I stumped when it comes to figuring a reason for the repetion though.

    Any ideas anyone?

    Thanks

    Pro
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  2. #2
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    Re: lovely result, but curious

    1^2+7^2=2(5)^2

    11^2\left(1^2+7^2\right)=2(11)^2\;5^2\Rightarrow\ 11^2+\left[11(7)\right]^2=2\left[11(5)\right]^2

    You can continue with

    111^2\left(1^2+7^2\right)=2(111)^2\;5^2

    and so on.
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  3. #3
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    Re: lovely result, but curious

    Thanks Archie,

    I knew there must be an answer, I just couldn't see it for looking, lol

    Pro
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  4. #4
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    Re: lovely result, but curious

    Hello, procyon!

    Whilst meandering through integers that fit the equation a^2+b^2\:=\:2c^2

    I came across this curious result
    . . \begin{array}{ccc}1^2+7^2 &=& 2(5^2) \\ 11^2+77^2 &=& 2(55^2) \\ 111^2+777^2 &=& 2(555^2) \\ \vdots && \vdots \end{array}

    I went as far as 16 digits and it still holds true.
    I stumped when it comes to figuring a reason for the pattern though.
    Any ideas anyone?

    \begin{array}{cc}\text{An integer consisting of }n\text{ 1's has the form:} & \dfrac{10^n-1}{9} \\ \\[-3mm] \text{An integer consisting of }n\text{ 7's has the form:} & 7\!\cdot\!\dfrac{10^n-1}{9} \\ \\[-3mm] \text{An integer consisting of }n\text{ 5's has the form:} & 5\!\cdot\!\dfrac{10^n-1}{9} \end{array}

    We wish to show that: . \left(\frac{10^n-1}{9}\right)^2 + \left(7\!\cdot\!\frac{10^n-1}{9}\right)^2 \;=\;2\left(5\!\cdot\!\frac{10^n-1}{9}\right)^2



    The left side is: . \left(\frac{10^n-1}{9}\right)^2 + 49\left(\frac{10^n-1}{9}\right)^2 \;=\;50\left(\frac{10^n-2}{9}\right)^2

    . . . . . . . . . . =\;2\cdot25\left(\frac{10^n-1}{9}\right)^2 \;=\; 2\left(5\!\cdot\!\frac{10^n-1}{9}\right)^2
    Q.E.D.

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  5. #5
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    Re: lovely result, but curious

    Hi Soroban,

    Very nice proof

    Thank you

    Pro
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  6. #6
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    Re: lovely result, but curious

    Quote Originally Posted by procyon View Post
    Thanks Archie,

    I knew there must be an answer, I just couldn't see it for looking, lol

    Pro
    All you have is "two sides are equal, so any common multiple of those two sides are equal"

    1^2+7^2=(2)5^2

    k^2\left(1^2+7^2\right)=2k^2\left(5^2\right)

    In this case, k^2=\left[(1.1)10^n\right]^2, for n=1, 2, 3,.....
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