Posers and Puzzles

Posers and Puzzles

  1. Joined
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    05 Jul '06 21:51
    Let's say the "repeat" of a positive whole number is the number obtained by writing it out twice in a row. So the repeat of 2006 would be 20062006.

    Is there a number whose repeat is a perfect square (ie. 1, 4, 9, 16, etc.)?
  2. Joined
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    05 Jul '06 23:49
    I think there is none. From 1-10000 they all had more than 20 (probably infinite) digits. Among the closest were:
    80838083 -> square-root: 8991.00011122233275662288
    99999999 -> square-root: 9999.99994999999987499999

    I'm trying to find a php-command to determine if a number is element of natural numbers. Then I can tell you right away instead of scrolling through endless pages of numbers 🙂

    However, no matter until what number I check, there will always be a larger number that might work. So you probably need a mathmatical proof why there isn't any number. But I don't know how to do that.
  3. Standard memberXanthosNZ
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    06 Jul '06 00:021 edit
    Originally posted by crazyblue
    I think there is none. From 1-10000 they all had more than 20 (probably infinite) digits. Among the closest were:
    80838083 -> square-root: 8991.00011122233275662288
    99999999 -> square-root: 9999.99994999999987499999

    I'm trying to find a php-command to determine if a number is element of natural numbers. Then I can tell you right away instead of scroll ...[text shortened]... robably need a mathmatical proof why there isn't any number. But I don't know how to do that.
    I also went for a computational approach rather than a proof one because I figured that it might give a quick answer.

    I was working in Matlab but I think I've got one up on you in terms of determining whether a number is a repeat.

    Say N is the number we are working with.
    floor(N/ceil(log(N))) is the (hopefully) repeating section. And then:
    if N - floor(N/ceil(log(N))) = ceil(log(N))*floor(N/ceil(log(N))) we have a repeating number.

    My code will only return something if it finds such a square which satisifies this condition. Checking up to 10 million returns nothing.
  4. Joined
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    06 Jul '06 00:221 edit
    Originally posted by XanthosNZ
    I also went for a computational approach rather than a proof one because I figured that it might give a quick answer.

    I was working in Matlab but I think I've got one up on you in terms of determining whether a number is a repeat.

    Say N is the number we are working with.
    floor(N/ceil(log(N))) is the (hopefully) repeating section. And then:
    if N - f ...[text shortened]... finds such a square which satisifies this condition. Checking up to 10 million returns nothing.
    i've almost got it, if all numbers of the form (10^n)+1 are square free it can't be done.
    if p^2|(10^n)+1 then LET r = (10^n +1)/p^2 since 2,3,5 don't divide 10^n+1 p^2 >=49 therefore number of digits in r is less than or equal to n keep multiplying r by 4 until you get an n-digit number call it X.
    repeat X is X*(10^n+1)= [(4^k)*r]*[r *(p^2)]=(2^k)^2*(r^2)*(p^2)

    if 10^n+1 is always square free then repeat of an n-digit number x
    is x*(10^n+1) if thats a square (10^n+1)|x but thats a contradiction because 10^n+1 {an (n+1)-digit num} > x {an n-digit num}

    EDIT: i'm using a|b to mean "a divides b"
  5. London
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    06 Jul '06 00:55
    Originally posted by aginis

    if 10^n+1 is always square free then repeat of an n-digit number x
    is x*(10^n+1) if thats a square (10^n+1)|x but thats a contradiction because 10^n+1 {an (n+1)-digit num} > x {an n-digit num}
    I don't follow you there . . . ?
  6. Joined
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    06 Jul '06 01:051 edit
    ok got it, 10 is a primitive root of {Z/(289)Z} to check that note that phi(289)=phi(17^2)=17^2-17=17*16
    therfore if 10^17 != 1mod 289 and 10^16 != 1mod289 then 10 is a primitive root. therefore exists k s.t. 10^k = (-1)mod289 =>
    10^k+1 = 0mod289 ergo 289|10^k+1 and 10^k+1 is not square free.

    namely k=phi(289)/2 = 17*8 = 136
    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
    588235294117647058823529411764705882352941176470588235294
    1176470588235294117648

    this was the smallest i could find
  7. Joined
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    06 Jul '06 01:47
    Originally posted by aginis
    ok got it, [...]

    n = 885813148788927335640138408304498269896193771626297577854
    671280276816608996539792387543252595155709342560553633217
    9930795847750865051904

    r = 941176470588235294117647058823529411764705882352941176470
    588235294117647058823529411764705882352941176470588235294
    1176470588235294117648

    this was the smallest i could find
    Are you really saying n is the square root of r here? I don't understand how you found that number, but I'm impressed 🙂
  8. Joined
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    06 Jul '06 01:50
    Originally posted by TommyC
    I don't follow you there . . . ?
    ok u want to repeat a number X with n digits
    step 1 add n zeroes to the end of the number (same as multiplying X by 10^n)
    step 2 add X to your result (X*10^n + X) = X * (10^n + 1)
    i want that to be a square so i try X = (10^n + 1) but that has one too many digits so i'm going to divide out a square (a square divided by a square is still a square) from X = (10^n + 1)
    in order to make it have less digits.

    so i need to find a square that divides (10^n + 1)
    so far i haven't fixed n so i can fiddle around with it using mods i figured out that 289|10^136+1 now X is too short so i multiply by 4 until i get a number of the right length (multiplying a square by a square is still a square) therfore my new X1*(10^136 + 1) is a square.
    but thats equal to repeat X1 (because X1 is of length 136)
    to find the root i just sqrt all my actions divide by 289 became divide by 17, and multiply by 256 became multiply by 16.
  9. Joined
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    06 Jul '06 01:511 edit
    Originally posted by crazyblue
    Are you really saying n is the square root of r here? I don't understand how you found that number, but I'm impressed 🙂
    no r is the square root of n repeated as in the number nn.
    i'm kinda impressed myself i didn't think i had it in me 😵
  10. Standard memberPalynka
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    06 Jul '06 01:57
    Originally posted by aginis
    ok u want to repeat a number X with n digits
    step 1 add n zeroes to the end of the number (same as multiplying X by 10^n)
    step 2 add X to your result (X*10^n + X) = X * (10^n + 1)
    i want that to be a square so i try X = (10^n + 1) but that has one too many digits so i'm going to divide out a square (a square divided by a square is still a square) from X = ...[text shortened]... t all my actions divide by 289 became divide by 17, and multiply by 256 became multiply by 16.
    Excellent reasoning.
  11. London
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    06 Jul '06 02:031 edit
    Originally posted by aginis
    ok u want to repeat a number X with n digits
    step 1 add n zeroes to the end of the number (same as multiplying X by 10^n)
    step 2 add X to your result (X*10^n + X) = X * (10^n + 1)
    i want that to be a square so i try X = (10^n + 1) but that has one too many digits so i'm going to divide out a square (a square divided by a square is still a square) from X = t all my actions divide by 289 became divide by 17, and multiply by 256 became multiply by 16.
    I understand the notation, but still don't follow your logic when it comes to repeat(x)=x(1 + 10^n). The only thing I could extract was that the unit part of x could not be 2, 3, 7 or 8, if there existed a natural number m st m^2 = repeat(x).

    Ho hum, I did only scrape a 2.1, many years ago now . . .

    Btw, congratulations on finding an answer.
  12. Joined
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    06 Jul '06 02:27
    Originally posted by TommyC
    I understand the notation, but still don't follow your logic when it comes to repeat(x)=x(1 + 10^n). The only thing I could extract was that the unit part of x could not be 2, 3, 7 or 8, if there existed a natural number m st m^2 = repeat(x).

    Ho hum, I did only scrape a 2.1, many years ago now . . .

    Btw, congratulations on finding an answer.
    more later 5 AM must sleeeeeeep
  13. London
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    06 Jul '06 02:36
    Originally posted by aginis
    more later 5 AM must sleeeeeeep
    That's ok, my curiousity has passed. Like Mephistopheles, I prefer negation to achievement. Thanks though.
  14. Joined
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    06 Jul '06 07:212 edits
    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 an one.
    The numbers of figures of 1(n)1 is always two more than n.
    Ex: When n=3 then 1(n)1 = 1(3)1 = 10001. 1(0) = 11.

    Okey here we go.
    I'd like to prove that RR is never a square.

    We first take R=2006 as an example. Then RR=20062006.
    To make R to RR we have to multiply R with 1(4)1. 2006*10001=20062006.
    That means 1(4) is a factor in 20062006.
    If we divide RR with 1(4)1 then we come back to R.
    If 1(4)1 is not by itself a square then 20062006 is not a square either.

    If we take any R with n figures, then RR is divisible with 1(n)1.
    And if 1(n)1 is not a square then RR can't be a square either.

    Can 1(n) ever be a square by itself?
    I say no. Here's the proof of that.

    1(0)1 * 1(0)1 = 11 * 11 = 121 = 1(0)2(0)1
    1(1)1 * 1(1)1 = 101 * 101 = 10201 = 1(1)2(1)1
    1(2)1 * 1(2)1 = 1001 * 1001 = 1002001 = 1(2)2(2)1
    and in general
    1(n)1 * 1(n)1 = 1(n)2(n)1.

    Since every 1(n)1 squared always has a two in the middle it can never be a square.

    And that proves that a repeat can never be a square.
  15. Joined
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    06 Jul '06 07:56
    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.
    ok first
    "If we divide RR with 1(4)1 then we come back to R.
    If 1(4)1 is not by itself a square then 20062006 is not a square either."
    this is logic is false if you multiply two non-squares {R and 1(n)1 } you can still get a square in return. Examples. 28*63 or 3*27 or in general (p*p*k)*(q*q*k).

    second your proof that 1(n)1 is never a square is incomplete, you only prove that the square root of a number of the form 1(n)1 is not of that form, what about other forms?

    I prefer this method:

    Let x^2 = 1(n)1
    3 does not divide 1(n)1 therefore 3 does not divide x
    (x-1)*(x+1)=x^2 - 1 = 1(n)0 = 10^k
    if 3 does not divide x, 3 divides either (x -1) or (x+1)
    but 3 does not divide 10^k = (2*5)^k therefore we have a contradiction.

    Essentially that is the same as trying to find solutions for the congruence x^2~2mod3 (legendre's symbol denotes a square remainder a mod prime p as (a/p)=1 and a non-square remainder a mod p as (a/p)=-1 (and 0 if a ~ 0mod p))
    since (1mod3)^2 = (1^2)mod3 = 1mod3
    (2mod3)^2 = 4mod3=1mod3 there are no solutions to x^2~2mod3
    since we are looking for x^2~1(n)1 mod 3~2mod3 there are no integer roots mod 3 but if "a" is a perfect square then it must have an integer root mod (p) for all p. therefore 1(n)1 is not a perfect square.

    examples
    1~1mod3
    4~1mod3
    9~0mod3
    16~1mod3
    25~1mod3
    ...
    437^2 = 190969 = 3*63656+1~1mod3

    finally see an example above for a number which when repeated is a square.
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