If $$f(x^{2} + f(y)) = xf(x) + y$$ for all non-negative integers x and y, then the value of $$[f(0)]^{2} + f(0)$$ equals________.

Putting $$x=0$$ in the given equation, we get the following relation,

$$f(f(y)) = y$$

Thus, the function produces the same output when it is iterated twice, once on $$y$$ and then on $$f(y)$$. Since $$f(x)=x$$ is a function that satisfies this condition, we can assume $$f(x)$$ to be equal to $$x$$ for every real number $$x$$. 

Therefore $$f(0)=0$$, and putting this value in the final equation, we get $$[f(0)]^2 + f(0) = 0+0= 0$$