Asked by Timothy
                The squares of three positive integers are in arithmetic progression, and the third integer is 12 greater than the first. Find the second integer.
            
            
        Answers
                    Answered by
            Hypocrites
            
    is there any more info to help us answwer
    
                    Answered by
            Timothy
            
    Nope, there is no additional information.
    
                    Answered by
            Hypocrites
            
    oh okay hold on
    
                    Answered by
            Reiny
            
    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 = +√(x^2 + 12x + 72) , so we are looking for perfect squares for x^2 + 12x + 72
let x = 1, y = √85 , not an integer
let x = 2, y = √100 , well that was lucky
so x = 2, y = 10 and z = 14
check:
their squares are : 4, 100, and 196, which is in AP
    
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 = +√(x^2 + 12x + 72) , so we are looking for perfect squares for x^2 + 12x + 72
let x = 1, y = √85 , not an integer
let x = 2, y = √100 , well that was lucky
so x = 2, y = 10 and z = 14
check:
their squares are : 4, 100, and 196, which is in AP
                    Answered by
            Anonymous
            
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