The sum of 3 numbers is 11, the product of the numbers is 24 and the sum of their squares is 53. What are the numbers, are they integers?
Ok well if we have the sum of these three things, we can find a cubic equation which can be solved.
Suppose our answers are a,b and c.
(a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab+ac+bc)
So (ab+ac+bc) = ((a+b+c)^2 - (a^2+b^2+c^2))/2
Now (x-a)(x-b)(x-c) = x^3 - (a+b+c)x^2 + (ab+ac+bc)x - abc
So our equation is
x^3 - (a+b+c)x^2 + ((a+b+c)^2-(a^2+b^2+c^2))/2 - abc = 0
In the case above we have
x^3 - 11x^2 + (121-53)/2x - 24 = x^3 - 11x^2 + 34x - 24
We can solve this using analytical methods to get roots = 1,4,6
x^2+y^2+z^2 = 53
(1,4,6) is the answer.
I just tried a bunch of values. I am not sure of how to solve a problem like this in general. I will think about it, and see if there is a method.