Find three consecutive integers such that the square of the sum of the smaller two is 144 more than the square of the largest.
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To solve this problem, let's assign variables to the three consecutive integers. We can call the smallest integer n, the middle integer n+1, and the largest integer n+2.
According to the information given, the square of the sum of the smaller two integers is 144 more than the square of the largest integer. Mathematically, we can represent this as:
(n + n+1)^2 = (n+2)^2 + 144
Now, let's simplify the equation and solve for n:
(n + n+1)^2 = (n+2)^2 + 144
(n + n+1)(n + n+1) = (n+2)(n+2) + 144
(n^2 + 2n + 1) = (n^2 + 4n + 4) + 144
n^2 + 2n + 1 = n^2 + 4n + 148
2n - 4n = 148 - 1
-2n = 147
n = -147/2
n = -73.5
However, since we are looking for consecutive integers, we cannot have a fraction as our smallest integer. Therefore, there is no solution to this problem using consecutive integers.
Alternatively, if you are allowed to consider consecutive even or consecutive odd integers, you can modify the approach accordingly.