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  • True and False with A Difference of Two Squares Dilemma

    Anonymous Christian
    Mar 28, 2019
    Math Olympiad, Math Olympiad Secondary, SASMO
    Comments Off on True and False with A Difference of Two Squares Dilemma

    Question in Image Below

     

     

     

     

    Solution

    22) Fact

     

    Perfect Squares can only have last digits 0, 1, 4, 5, 6, 9

     

    Perfect Square CANNOT have these last digits: 2, 3, 7, 8

    If (i) and (ii) were true, a number with last digit 2 is a perfect square, impossible!

    If (ii) and (iii) are true, a number with last digit 1 – 48 has last digit 3 is a perfect square, contradiction again!

    The only possibility is that statement (i) and (iii) must both be true

    Statement (ii) is false

    A + 41 = x^2   and A – 48 = y^2

     

    Difference

     

    x^2 – y^2 = 89 , x > y

     

    (x – y)(x + y) = 89 and 89 is a prime number

     

    (x – y) = smaller factor = 1

     

    and

     

    (x + y) = 89

     

    Adding these:

     

    (x – y) + (x + y) = 1 + 89

     

    2x = 90

     

    x = 45

     

    Put back in (x – y) = 1

     

    45 – y = 1

     

    y = 44

     

     

    Finally

     

    A + 41 = x^2

     

    Becomes

     

    A + 41 = 45^2

     

    A = 1984

    Check

     

    (i) A + 41 is a Perfect Square

     

    1984 + 41 = 2025 = 45^2 (True)

     

     

     

     

    (ii) Last digit of A = 1984 is 1 (False)

     

     

     

     

     

    (iii) A – 48 is a Perfect Square

     

    1984 – 48 = 1936 = 44^2 (True)

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