The squares of three positive integers are in arithmetic progression, and the third integer is 12 greater than the first. Find the second integer.

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  1. is there any more info to help us answwer

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  2. Nope, there is no additional information.

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  3. oh okay hold on

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  4. Let the 3 positive integers be x, y and z
    but z = x+12

    we have:
    y^2 - x^2 = (x+12)^2 - y^2
    2y^2 = 2x^2 + 24x + 144
    y^2 = x^2 + 12x + 72
    y = +āˆš(x^2 + 12x + 72) , so we are looking for perfect squares for x^2 + 12x + 72
    let x = 1, y = āˆš85 , not an integer
    let x = 2, y = āˆš100 , well that was lucky

    so x = 2, y = 10 and z = 14

    their squares are : 4, 100, and 196, which is in AP

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