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Math Help - modified pythagorus theorem!

  1. #1
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    Post modified pythagorus theorem!

    Hello,

    I need a little help in figuring this weird problem...

    --we know that (2+3)^2 is not equal to 2^2 and 3^2...
    however... (2+3)^2 = (2+1)^2 + (3+1)^2!

    so determine, all values for m and n such that---
    (m+n)^2 = (m+1)^2 + (n+1)^2

    Thanks!
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  2. #2
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    Quote Originally Posted by Vedicmaths View Post
    Hello,

    I need a little help in figuring this weird problem...

    --we know that (2+3)^2 is not equal to 2^2 and 3^2...
    however... (2+3)^2 = (2+1)^2 + (3+1)^2!

    so determine, all values for m and n such that---
    (m+n)^2 = (m+1)^2 + (n+1)^2

    Thanks!
    (m+n)^2 = (m+1)^2 + (n+1)^2
    expand & simplify results in
     mn = m+n+1
    If the product of any two numbers is equal to the sum of the two numbers plus 1, then they work.

    Is that what you had in mind?
    2,3 is a solution that you have given.
    It appears that is the only solution in natural numbers.
    .
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  3. #3
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    Quote Originally Posted by Vedicmaths View Post
    Hello,

    I need a little help in figuring this weird problem...

    --we know that (2+3)^2 is not equal to 2^2 and 3^2...
    however... (2+3)^2 = (2+1)^2 + (3+1)^2!

    so determine, all values for m and n such that---
    (m+n)^2 = (m+1)^2 + (n+1)^2

    Thanks!
    Hmmm....isn't this trivial? Just develop parentheses in both sides!

    m^2 + 2mn + n^2 = m^2 + 2m + 1 + n^2 + 2n + 1 ==> 2mn = 2(m+n+1) <==> mn - m - n = 1 <==> (m-1)(n-1) = 2, so we don't have that many options for m, n , do we?

    Tonio
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  4. #4
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    Post

    thanks for the response!

    yeah I also simplified that and ended up getting (m-1)(n-1) = 2...that means m and n can only have 2 or 3 values respectively.
    I was just making it too complicated..I thought it would be a little bit harder that it seems!

    Hence, m =2 or 3 and n = 3 or 2...right?


    Thanks a lot!
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  5. #5
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    Quote Originally Posted by Vedicmaths View Post
    thanks for the response!

    yeah I also simplified that and ended up getting (m-1)(n-1) = 2...that means m and n can only have 2 or 3 values respectively.
    I was just making it too complicated..I thought it would be a little bit harder that it seems!

    Hence, m =2 or 3 and n = 3 or 2...right?


    Thanks a lot!

    Yep. Since the equation (m-1)(n-1) = 2 is symmetric wrt m, n there's only one solution in the naturals: m = 3, n = 2.
    If you allow integers in general there's another solution: m = 0, n = -1

    Tonio
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