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    The repeat of a positive integer is obtained by writing it twice in a row (so for example the repeat of 213 is 213213). Is there a positive integer whose repeat is a square number?


    i said call the repeat number abcabc and assume the result is true such that abcabc=n^2 for some n of the set of natural numbers.

    Then abcabc=10*101[100a+10b+c]=n^2 (writing out the sum of each digit and factorising)

    Then for LHS to be perfect square:P 100a+10b+c=10*101 as 1010 is not a square and is not decomposed of squares since 10=5*2.

    But we also must restrict 0<a<9 and same for b and c. So the max value of 100a+10b+c is 100(9)+10(9)+9=999 (in other words it is at most a 3 digit number)

    So as 1010>999 none such repeat numbers exist as perfect squares


    Is this argument ok? infact i have no idea if the conclusion is right as no answer was published alongside the question.

    (after typing this up and rereading the question i have realised that although the example was a 3 digit number, the number could be as long as you want: i think if this proof is right i could extend it and you would always have the maximum 99...999< SOMENUMBER needed. )

    What do you think?
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    (Original post by newblood)
    The repeat of a positive integer is obtained by writing it twice in a row (so for example the repeat of 213 is 213213). Is there a positive integer whose repeat is a square number?


    i said call the repeat number abcabc and assume the result is true such that abcabc=n^2 for some n of the set of natural numbers.

    Then abcabc=10*101[100a+10b+c]=n^2 (writing out the sum of each digit and factorising)

    abcabc = 1001 * [100a+10b+c]=n^2
    And 1001 factorises as 7,11,13.

    Then for LHS to be perfect square:P 100a+10b+c=10*101 as 1010 is not a square and is not decomposed of squares since 10=5*2.
    At this stage your requirement is that 1001 (corrected) divides 100a+10b+c and that the quotient is a square: Since all the prime factors of 1001 have a multiplicity of 1.

    The rest of your argument follows, and works for a 3-digit number.

    I don't have a full answer for the question, so can't help further. You may need a different line of attack for a general number.
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    (Original post by ghostwalker)
    abcabc = 1001 * [100a+10b+c]=n^2
    And 1001 factorises as 7,11,13.



    At this stage your requirement is that 1001 (corrected) divides 100a+10b+c and that the quotient is a square: Since all the prime factors of 1001 have a multiplicity of 1.

    The rest of your argument follows, and works for a 3-digit number.

    I don't have a full answer for the question, so can't help further. You may need a different line of attack for a general number.
    ah i must have made a slight arithmetic error there,

    thank you for you for your comments btw, much appreciated
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    This is a Cambridge example sheet question, the tag is hardly appropriate...

    (Original post by newblood)
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    I don't have time to read your working, but the conclusion is incorrect. Very strong hint: n^2\,\big|\,(n-1)^n+1 if n is odd.
 
 
 
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