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Find all the ordered pairs of integers such that x^2 - y^2 = 140

Any help or explanations would be wonderful!

  • math -

    the hyperbola x^2 - y^2 = 140
    can be factored to
    (x+y)(x-y) = 140


    140 = 2x5x2x7
    or in pairs:
    14x10 or 28x5 or 2x70, or 35x4

    so we are looking for any two numbers x and y
    so that their sum x their difference is 140, using only the numbers above
    e.g x+y=28
    x-y = 5
    add them: 2x = 33 ---> x not an integer

    how about:
    x+y=70
    x-y=2
    add them: 2x = 72
    x = 36, then y = 34
    then (36+34)(36-34) = 140

    (36,34), (36,-34), (-36,34) , (-36,-34)

    a very limited number of cases.
    how about (12+2)(12-2) ?

    so (12,2), (-12,2), (12,-2) and (-12,-2)

  • math -

    I left out
    140 = 7x20
    but x+y=20
    x-y=7 has no integer solution, since 2x would have to be even.

  • math -

    Thank you very much!

  • math -

    Since
    x²-y²
    =(x+y)(x-y)
    we are looking for two integers (x+y) and (x-y) that have a product of 140, and such that x and y are integers.

    We can start by enumerating the factors of 140:
    140*1
    70*2
    35*4
    28*5
    20*7
    14*10

    Since x and y are both integers, and (x+y) and (x-y) must also be integers. This implies that (x+y) and (x-y) must be either both odd or both even (for proof, see end of post).
    The only pairs that meet this criterion are 14,10 and 70,2, which when solved give (x,y)=(12,2), or (36,34).

    Given x, y ∈ℤ, proof that (x+y) and (x-y) are either both even or both odd.
    We have to address two cases:
    If (x-y) is even, then
    x+y
    =(x-y)+2y
    =2k+2y
    =2(k+y)... so x+y is also even.
    If (x-y) is odd, then
    x+y
    =(x-y)+2y
    =(2k+1)+2y
    =2(k+y)+1....so x+y is also odd.
    QED

  • math-correction -

    Thanks Reiny, I left out combinations of cases where one or both of x and y is negative.

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