1. Standard memberXanthosNZ
    Cancerous Bus Crash
    p^2.sin(phi)
    Joined
    06 Sep '04
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    25076
    06 Jul '06 08:44
    Originally posted by FabianFnas
    My solution is something like this:

    Def: R is a positive number not started with a zero. Then RR is R repeated twice, we call it a 'repeat'. The number of figures in RR is always twice the number of R.
    Ex: If R = 2006 then RR = 20062006.

    Def: 1(n)1, when n is a non-negative number, is a number starting with a one followed by n zeroes, ending with a ...[text shortened]... he middle it can never be a square.

    And that proves that a repeat can never be a square.
    Why do people do this? An example has been posted so unless you can find a mistake in the example then your proof that no example exists must be incorrect.
  2. Joined
    11 Nov '05
    Moves
    43938
    06 Jul '06 08:45
    Maybe I was a little vague here and there. Good that you are abservant.
    It was crystal clear when I formulated the 'proof'.

    You wrote:
    "437^2 = 190969 = 3*63656+1~1mod3
    finally see an example above for a number which when repeated is a square."


    Please, again, show me the number, R,when repeated, RR, is a square.
  3. Joined
    11 Jun '06
    Moves
    3516
    06 Jul '06 08:551 edit
    Originally posted by aginis
    so let n = 256*(10^136 + 1)/289 (has 136 digits)
    let r = 16* (10^136 + 1)/17 (also has 136 digits)

    r^2 = (repeat n) (has 272 digits)
    n = 885813148788927335640138408304498269896193771626297577854
    671280276816608996539792387543252595155709342560553633217
    9930795847750865051904
    r = 941176470588235294117647058823529411764705882352941176470
    5882352941176470588235294117647058823529411764705882352941176470588235294117648
    nn = 885813148788927335640138408304498269896193771626297577854
    671280276816608996539792387543252595155709342560553633217
    993079584775086505190488581314878892733564013840830449826
    989619377162629757785467128027681660899653979238754325259
    51557093425605536332179930795847750865051904 =

    941176470588235294117647058823529411764705882352941176470
    588235294117647058823529411764705882352941176470588235294
    1176470588235294117648^2
  4. Joined
    20 Feb '06
    Moves
    8407
    06 Jul '06 09:231 edit
    aginis is correct here, well done!

    But there's a smaller example than the one given above.

    It all depends on whether or not we can find a square factor of 10^n + 1 for some n.

    In fact

    10^11 + 1 = 11^2 x 23 x 4093 x 8779

    and so (10^11 + 1) x 23 x 4093 x 8779 is square. Unfortunately this square is not the repeat of 23 x 4093 x 8779, which has too few digits, but we can simply multiply it by 4^2 to get it in the right range.

    In fact 23 x 4093 x 8779 x 4^2 = 13223140496 and

    1322314049613223140496 = 36363636364^2

    where of course 36363636364 = 11 x 23 x 4093 x 8779 x 4.
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