Results 1 to 4 of 4

Math Help - Determining pairs (m,n) for...?

  1. #1
    Super Member fardeen_gen's Avatar
    Joined
    Jun 2008
    Posts
    539

    Determining pairs (m,n) for...?

    Determine all pairs (m,n) of positive integers m, n for which 2^m + 3^n is a perfect square.

    P.S. Is there any rigid approach for such problems? Do they require more of an intuitive approach?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Lord of certain Rings
    Isomorphism's Avatar
    Joined
    Dec 2007
    From
    IISc, Bangalore
    Posts
    1,465
    Thanks
    6
    Quote Originally Posted by fardeen_gen View Post
    Determine all pairs (m,n) of positive integers m, n for which 2^m + 3^n is a perfect square.

    P.S. Is there any rigid approach for such problems? Do they require more of an intuitive approach?

    Its unfortunate that I have a long messy,sloppy approach... I have a hunch that there is an elegant method using clever number theoretic arguments... but I dont know it.

    u^2 = 2^m + 3^n \mod 3 = (-1)^m \mod 3

    But modulo 3, squares are always 0 or 1, thereby forcing m to be even.

    Now lets call m = 2k

    3^n = u^2 - 2^{2k} = (u - 2^k)(u+2^k)

    The factors of power of three are all lesser powers of three. Thus for some r, we get 2 equations:

    u - 2^k = 3^r
    u + 2^k = 3^{n-r}

    Now note that this implies 2^{k+1} = 3^{n-r} - 3^r = 3^r(3^{n-2r} - 1)

    r > 0 is not possible because then the RHS is divisible by 3 and the LHS is not.

    Thus r = 0. This forces the equations to read:

    2^{k+1} =3^n - 1

    Reading modulo 4 now, 0 =(-1)^n - 1 \mod 4. This forces n to be even.

    Let n = 2l, and repeat the same old process,

    2^{k+1} = 3^{2l} - 1

    The factors of power of 2 are all lesser powers of two. Thus for some r, we get 2 equations:

    3^l  - 1 = 2^r
    3^l + 1 = 2^{k+1-r}

    2^{k+1-r} - 2^r = 2-----------------------------(**)

    The above equation clearly forces r = 1,k+1-r = 2 hence k = 2.

    Now since r=1, 3^l = 2^r + 1 = 3 \Rightarrow l=1

    Thus \boxed{n = 2l = 2} and \boxed{m = 2k = 4} are the only solutions.

    Check: 2^4 + 3^2 = 25 = 5^2








    --------------------------------------------------------------------------------------------------------------------------------------------
    (**) 2^x - 2^y = 2 \Rightarrow x=2,y=1

    2^x - 2^y = 2 \Rightarrow 2^{x-1} - 2^{y-1} = 1

    There are now various ways to solve it. One simple thing we can observe is that for x,y > 1, RHS mod 2 is 1, but LHS mod 2 is 0. So at least one of x and y is 1. Clearly x = 1 is not possible. The only remaining condition: y=1 forces x=2.

    Thus 2^x - 2^y = 2 \Rightarrow x=2,y=1


    Last edited by Isomorphism; July 14th 2008 at 07:18 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    your solution is actually pretty good, but there's one very little mistake (probably typo), which can be easily fixed:

    Quote Originally Posted by Isomorphism View Post

    2^{k+1} =3^n - 1

    Reading modulo 2 now, 0 =(-1)^n - 1 \mod 2. This forces n to be even.
    modulo 2 both sides are 0 for all values of n, even or odd. but since k > 0, the LHS is 0 modulo 4, which forces n to be even.
    Last edited by NonCommAlg; July 14th 2008 at 06:46 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Lord of certain Rings
    Isomorphism's Avatar
    Joined
    Dec 2007
    From
    IISc, Bangalore
    Posts
    1,465
    Thanks
    6
    Quote Originally Posted by NonCommAlg View Post
    your solution is actually pretty good, but there's one very little mistake (probably typo), which can be easily fixed:

    modulo 2 both sides are 0 for all values of n, even or odd. but since k > 0, the LHS is 0 modulo 4, which forces n to be even.
    Thanks for pointing that. It is a typo that I made while copying from my rough work.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. How many different ordered pairs?
    Posted in the Geometry Forum
    Replies: 4
    Last Post: November 12th 2008, 04:37 PM
  2. number of pairs
    Posted in the Algebra Forum
    Replies: 2
    Last Post: June 18th 2008, 08:41 AM
  3. Ordered Pairs (x,y)
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: June 11th 2007, 04:04 PM
  4. Two Ordered Pairs
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: June 9th 2007, 07:34 AM
  5. Pairs of (x,y)
    Posted in the Math Topics Forum
    Replies: 4
    Last Post: May 28th 2007, 07:11 AM

Search Tags


/mathhelpforum @mathhelpforum