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Pure mathematics problem

  1. Oct 10, 2009 #1
    1. The problem statement, all variables and given/known data
    Prove that 10^n + 1 cannot be expressed in the form a*a*c, where n,a,c are positive integers.


    2. Relevant equations
    By considering the reminder of 10^n + 1 when it is devided by 3, I arrived at the conclusion that:

    a*a = 3z +1;
    c = 3v + 2

    for some positive integers z,v.

    Here I am stuck, I have no idea how to proceed


    3. The attempt at a solution
     
  2. jcsd
  3. Oct 10, 2009 #2

    Mark44

    Staff: Mentor

    How did you arrive at this conclusion? You don't show what you got when you divided 10n + 1 by 3.
     
  4. Oct 11, 2009 #3
    Let's consider 3 possible cases:

    1) a = 3k. This case is impossible, since from equality 10^n + 1 = a*a*c we notice that the remainder, when LHS is divided by 3, is always 2, and the remainder, when RHS is divided by 3, is 0;

    2) a = 3k + 1. Then a*a = 9k*k + 6k + 1 - the remainder, when a is divided by 3, is 1

    3) a = 3k + 2. Then a*a = 9k*k + 12k + 4 - the remainder, when a is divided by 3, is 1

    Hence, a*a can be expressed as a*a = 3z + 1

    Now let's consider 3 possible cases for c:

    1) c = 3k - this case is impossible due to the same reason explained in case a = 3k.

    2) c= 3k + 1. Then:

    a*a*c = (3z + 1)(3k + 1) = 9kz + 3z + 3k + 1. Dividing this by 3 gives remainder 1, while dividing 10^n + 1 by 3 gives remainder 2

    3) c= 3k + 2. Then:

    a*a*c = (3z + 1)(3k + 2)= 9kz + 6z + 3k + 2. Dividing this by 3 gives remainder 2, and dividing 10^n + 1 by 3 gives remainder 2.

    Hence, c= 3k + 2.



    (The real problem I want to solve is formulated like that:

    The repeat of a natural number is obtained by writing it twice in a row (for example, the repeat of 356 is 356356). Is there any number whose repeat is a perfecr square.

    Let's denote the n-digit number by b. Then we have:

    b*10^n + b = c*c;
    b(10^n + 1)= c*c

    I have made a little investigation for 10^n + 1:

    n
    1 10^n + 1=11 - prime number
    2 10^n + 1=101 - prime number
    3 10^n + 1=1001=7*11*13
    4 10^n + 1=10001=73*137
    5 10^n + 1=100001=11*9091
    6 10^n + 1=1000001=101*9901
    7 10^n + 1=10000001=11*909091
    8 10^n + 1=100000001=17*5882353
    9 10^n + 1=1000000001= 7*11*13*19*52579

    Form this, we notice that each prime factor of 10^n + 1 occurs only once. Hence, I have come to the sketch of the proof that no such number whose repeat is a perfecr square exists:

    1)10^n + 1 is a prime number.
    Then it must be a= e*e(10^n + 1) -> a is a n+1 digit number -> contradiction

    2) 10^n + 1 is not a prime number

    Each prime factor of 10^n + 1 occur only once (as "deduced form examples above")
    I try to prove last statement by contradiction. Let's assume that 10^n + 1 = a*a*c (a*a means that some prime factors occur more that once). The work for this part was described above.

    Hence, if this last unproven statement is true, then a = e*e(10^n + 1) -> a is a n+1 digit number -> contradiction.

    )
     
  5. Dec 11, 2009 #4
    While this may be an old thread I stumbled across and felt I should add some input.

    This statement cannot be proven as it is currently written.

    Ex. Assign c the value 10^n + 1. Divide LHS by c. This implies that a^2 = 1 which implies a = 1. Therefore for all n 10^n + 1 can be written as a*a*c.

    If it was meant that a and c must be strictly less than 10^n + 1 then the statement can be much more easily proven then the route you were taking.

    Start by dividing LHS by c. Then it must be that ( 10^n + 1 ) / c is a perfect square since a is a positive integer. This implies that [tex]\sqrt{ 10^n / c + 1 / c }[/tex] is also an integer. All thats left for you to do is show that for every c < 10^n + 1 , that [tex]\sqrt{ 10^n / c + 1 / c }[/tex] is never an integer value.

    Hint: it would suffice to prove that the expression inside the brackets is itself never an integer value for all allowed values of c.
     
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