1. This site uses cookies. By continuing to use this site, you are agreeing to our use of cookies. Learn More.
  2. Hi Guest, welcome to the TES Community!

    Connect with like-minded education professionals and have your say on the issues that matter to you.

    Don't forget to look at the how to guide.

    Dismiss Notice

Proposed Problem

Discussion in 'Mathematics' started by AVDBA, Sep 17, 2019.

  1. AVDBA

    AVDBA New commenter

    Consider the equation:

    (x + y)k = xy,

    where x and y are integers. Given that k is a prime number, find the solution set {(x, y)}.
     
  2. primenumbers

    primenumbers New commenter

    This is a standard problem in Junior and Intermediate Olympiad papers. Are you asking for help with your hw or want to discuss it?
     
  3. AVDBA

    AVDBA New commenter

    Thanks primenumbers. I am providing these latter proposed problems, since I think these are two examples that explore a process of "working with the concepts". For these two cases, they are related to interconnecting concepts regarding factorability. They are examples of "exploring the concepts throughout". The related concepts are, for these specifically, their tags. This problem here, specifically, requires factorization respecting the stated concepts, as well as the analysis to impose the condition that the solution must obey the necessity required by the set as stated by the problem's statement. Importantly, the solution requires "testing controlled values", a concept that is related to the scientific method. I have not stated these problems from Olympiad papers. They are of type within a sample of problems that are explored here for entering universities.
     
  4. AVDBA

    AVDBA New commenter

    Hints for working throughout the solution:

    1 - To rewrite the considered equation by isolating (LHS), e.g., x, keeping in mind the meaning of k at RHS, and the concept of divisibility.

    4 - To test all values that fall into the required set (controlled test).
     
  5. primenumbers

    primenumbers New commenter

    Please ignore me since I am not sure what you are trying to tell me.

    I can only solve Maths problems by applying what I have learnt. If I fail, I will do a bit more research, learn a few more things and have another go. No idea about any science behind it.
     
  6. AVDBA

    AVDBA New commenter

    Thanks primenumbers. The hints were intended to guide through a solution. The controlled set is a reference to the necessity of testing values. The necessity of making observations to infer the validity of the predictions.

    The considered equation may be rewritten as:

    x = (ky)/(y - k)

    Since x is an integer, we have the following cases:

    * (y - k) = 1 or (y - k) = -1. This case leads to y = (k + 1) or y = (k - 1),
    and respectively to x = k(k + 1) or x = k(1 - k). The set {(x,y)} for this
    case reads: {(k(k + 1), k + 1); (k(1 - k), k - 1)}

    * y is a a multiple of (y - k), excluding the case previously analyzed -
    with (y - k) not belonging to the set {-1, 1}. This implies that y must
    be proportional to (y - k), i.e., y = m(y - k), where m is an integer,
    implying that x = mk, which requires that y = (kx)/(x - k). Comparing this
    latter equation with the original x = (ky)/(y - k), one infers that the
    solution set for this case is the same one that is obtained by
    interchanging y and x, i.e., being the set {(x,y)} a solution, {(y,x)} is
    the solution for the case being analyzed here [y being a multiple
    of (y - k)].

    * k is a multiple of (y - k), excluding the case previously analyzed -
    with (y - k) not belonging to the set {-1, 1}. But, being k a prime number,
    k must satisfy for this case k = (y - k) or k = (k - y), from which, respectively,
    x = y or x = -y. Hence, for k = (y - k): y = k + k; for k = (k - y): y = k - k.
    These latter results requires, respectively, x = y = k + k or x = -y = -(k - k).
    The set {(x,y)} for this case reads: {(2k, 2k); (0, 0)}.

    The solution set then reads:
    {(x,y)} =
    {
    (k(k + 1), k + 1);
    (k(1 - k), k - 1);
    (2k, 2k);
    (0, 0);
    (k + 1, k(k + 1));
    (k - 1, k(1 - k))
    }

    The elements (k + 1, k(k + 1)) and (k - 1, k(1 - k)) were obtained
    by interchanging x a y, as required by the second case previously
    analyzed. Interchanging x and y in the elements (0, 0) and (2k, 2k)
    does not generate new members for the solution set {(x,y)}.
     

Share This Page