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.
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
Answered by Anonymous
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