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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.
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
Actually I was looking for a general answer. The trial and error method is a no-brainer.
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