The squares of three positive integers are in arithmetic progression, and the third integer is 12 greater than the first. Find the second integer.
5 answers
is there any more info to help us answwer
Nope, there is no additional information.
oh okay hold on
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
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1 answer