Results 1 to 5 of 5

Math Help - given positive integers a and b such that a|b^2, b^2|a^3, a^3|b^4,...

  1. #1
    Newbie
    Joined
    Jan 2010
    Posts
    13

    given positive integers a and b such that a|b^2, b^2|a^3, a^3|b^4,...

    The problem below seems fairly easy... but I just can't manage to prove it!

    PROBLEM: Given posistive integers a and b such that a|b^2, b^2|a^3, a^3|b^4, b^4|a^5, ... , prove that a = b.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by swallenberg View Post
    The problem below seems fairly easy... but I just can't manage to prove it!

    PROBLEM: Given posistive integers a and b such that a|b^2, b^2|a^3, a^3|b^4, b^4|a^5, ... , prove that a = b.

    a\mid b^2\Longrightarrow b^2=xa\,,\,b^2\mid a^3\Longrightarrow a^3=yb^2\Longrightarrow this already tells us that every prime that divides a divides b and every prime that divides b divides a, so a,b are divisible by exactly the same primes. Well, now just prove that if a prime divides a to some power, then at that same power that primes divides b, and the other way around.

    Tonio
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jan 2010
    Posts
    13
    Quote Originally Posted by tonio View Post
    Well, now just prove that if a prime divides a to some power, then at that same power that primes divides b, and the other way around.

    Tonio

    But how? I came this far when I tried the first time... but I just can't seem to cross the finish line.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    quote=swallenberg;446066]But how? I came this far when I tried the first time... but I just can't seem to cross the finish line.[/quote]


    Suppose p^\alpha\mid a\,,\,\,p^\beta\mid b and these powers are maximal in each case, i.e. p^{\alpha+1}\nmid a\,,\,\,p^{\beta+1}\nmid b, and suppose \alpha\leq \beta , then:

    a\mid b^2\Longrightarrow b^2=xa . But p^{2\beta}\mid b^2\Longrightarrow p^{2\beta}\mid xa\Longrightarrow p^{2\beta-\alpha}\mid x . Continue from here comparing odd powers of a with even powers of b.

    Tonio
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Jan 2010
    Posts
    13
    I finally realized how to do it.

    (2n-1)alfa
    ≤ 2nbeta ≤ (2n+1)alfa, which is the same as (2n-1)/2n ≤ beta/alfa ≤ (2n+1)/2n.

    Now if we let n tend to inf. we'll see that 1
    beta/alfa ≤ 1 which means that alfa = beta.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. if x and n are positive integers....
    Posted in the Math Topics Forum
    Replies: 2
    Last Post: November 7th 2009, 12:07 AM
  2. Positive Integers
    Posted in the Math Topics Forum
    Replies: 2
    Last Post: May 31st 2007, 02:48 PM
  3. Two Positive Integers....
    Posted in the Algebra Forum
    Replies: 6
    Last Post: May 20th 2007, 07:50 PM
  4. Sum of Positive Integers
    Posted in the Algebra Forum
    Replies: 4
    Last Post: March 31st 2007, 05:34 AM
  5. Positive Integers
    Posted in the Algebra Forum
    Replies: 1
    Last Post: January 28th 2007, 06:25 PM

Search Tags


/mathhelpforum @mathhelpforum