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Math Help - Find all pairs of positive integers a and b such that...

  1. #1
    Junior Member
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    Find all pairs of positive integers a and b such that...

    Once again I am stuck on a question. Thanks as always.

    Find all pairs of positive integers a and b such that...

    a^2 + b^2 - 7 = ab

    I've done this:
    a^2 - ab - (7-b^2) = 0

    But after that?

    I've also found two solutions:a=2, b=3 and a=3 and b=2.
    But i lack a solid algebraic proof.
    Thanks!
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  2. #2
    MHF Contributor
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    You have the idea but you did not pursue it.

    After your
    a^2 -ab -(7 -b^2) = 0 ----(i)
    You should have seen that that is a quadratic equation in "a" and so you could have used the quadratic formula to solve for "a" in terms of "b".

    Let me continue it.

    Suppose x = a, then (i) becomes
    x^2 -bx -(7 -b^2) = 0 ----see, is it familiar now?

    Back to (i),
    a^2 -ab -(7 -b^2) = 0 ----(i)
    Using the quadratic formula,
    a = {-(-b) +,-sqrt[(-b)^2 -4(-(7 -b^2)]} / (2*1)
    a = {b +,-sqrt[b^2 +4(7 -b^2)]} / 2
    a = {b +,-sqrt[b^2 +28 -4b^2]} / 2
    a = {b +,-sqrt[28 -3b^2]} /2 --------(ii)

    (ii) is "a" in terms of "b". So we assign values to "b" to find the corresponding "a" values.
    The question seeks positive integers only, so,

    When b = 1,
    a = {1 +,-sqrt[28 -3(1^2)]} / 2
    a = {1 +,-sqrt[25]} / 2
    a = {1 +,-5]} / 2
    a = 6/2 = 3
    Hence, (3,1) -------in the form (a,b) ordered pair.

    When b = 2,
    a = {2 +,-sqrt[28 -3(2^2)]} / 2
    a = {2 +,-sqrt[16]} / 2
    a = {2 +,-4]} / 2
    a = 6/2 = 3
    Hence, (3,2) -------***

    When b = 3,
    a = {3 +,-sqrt[28 -3(3^2)]} / 2
    a = {3 +,-sqrt[1]} / 2
    a = {3 +,-1]} / 2
    a = 4/2 = 2
    Hence, (2,3) -------***

    When b = 4,
    a = {4 +,-sqrt[28 -3(4^2)]} / 2
    a = {4 +,-sqrt[-20]} / 2
    Cannot be anymore, because sqrt(-20) is not real number.

    So, we found 3 pairs so far:
    (2,3), (3,1), and (3,2)

    ----------------------
    Analyzing the original a^2 + b^2 - 7 = ab again, if we see it in "b" way, then,
    b^2 -ab -(7 -a^2) = 0 ---(iii)

    Do you see any resemblance between (i) and (iii)?

    Yes, in (iii), "b" and "a" have reversed roles as those in (i).
    So if we solve for "b" in terms of "a" by using the quadratic formula, we will arrive to
    a=1, b=3 ----(1,3)
    a=2, b=3 ----(2,3) <---same or found already in (i)
    a=3, b=2 ----(3,2) <---already found in (i)

    ------------
    Therefore, the possible all pairs of positive "a" and "b" are:
    (1,3), (2,3), (3,1), and (3,2) --------answer.
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  3. #3
    Junior Member
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    Thanks very much once again ticbol! Very detailed explanation.
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