The number of points of discontinuity of the function $$f(x) = \left[\frac{x^2}{2}\right] - \left[\sqrt{x}\right], x \in [0, 4]$$, where $$[\cdot]$$ denotes the greatest integer function is ______.

The greatest-integer (floor) function $$[t]$$ is continuous at every real number except the integers, where it possesses a jump of magnitude $$1$$.
Hence the difference $$f(x)=\left[\frac{x^{2}}{2}\right]-\left[\sqrt{x}\right]$$ can be discontinuous only at those $$x$$ where at least one of the two quantities $$\dfrac{x^{2}}{2}$$ or $$\sqrt{x}$$ is an integer.

Step 1 : Locate points where $$\dfrac{x^{2}}{2}$$ is an integer.
Let $$\dfrac{x^{2}}{2}=k$$, where $$k\in\mathbb{Z}$$ and $$0\le x\le4$$.
Then $$x=\sqrt{2k}$$.
For $$x\in[0,4]$$ we need $$0\le\sqrt{2k}\le4\;\Longrightarrow\;0\le k\le8$$.
Thus the admissible integers are $$k=0,1,2,3,4,5,6,7,8$$, giving

$$x=0,\;\sqrt2,\;2,\;\sqrt6,\;\sqrt8,\;\sqrt{10},\;\sqrt{12},\;\sqrt{14},\;4$$.

Step 2 : Locate points where $$\sqrt{x}$$ is an integer.
Let $$\sqrt{x}=m$$ with $$m\in\mathbb{Z},\;0\le x\le4$$.
Then $$x=m^{2}$$ and, since $$0\le m\le2$$, we have

$$x=0,\;1,\;4$$.

Step 3 : Form the combined candidate set.
Taking the union of the two lists yields the following distinct points inside $$[0,4]$$:

$$0,\;1,\;\sqrt2,\;2,\;\sqrt6,\;\sqrt8,\;\sqrt{10},\;\sqrt{12},\;\sqrt{14},\;4.$$\

Step 4 : Check continuity at each candidate.

1. $$x=0$$ (left end-point).  For $$x\to0^{+}$$,
$$\left[\dfrac{x^{2}}{2}\right]=0,\quad\left[\sqrt{x}\right]=0\;\Longrightarrow\;f(x)=0.$$
At $$x=0$$ itself, $$f(0)=0-0=0$$, matching the right-hand limit. Continuous.

2. $$x=1$$.
Near $$1^{-}$$: $$\left[\dfrac{x^{2}}{2}\right]=0,\;\left[\sqrt{x}\right]=0\;\Rightarrow\;f=0.$$}
Near $$1^{+}$$: $$\left[\dfrac{x^{2}}{2}\right]=0,\;\left[\sqrt{x}\right]=1\;\Rightarrow\;f=-1.$$}
Jump $$\neq0$$ ⇒ discontinuous.

3. $$x=\sqrt2$$.
Here $$\dfrac{x^{2}}{2}=1$$ (integer) while $$\sqrt{x}\approx1.189\;(non\;integer).$$
Thus $$\left[\dfrac{x^{2}}{2}\right]$$ jumps from $$0$$ to $$1$$, whereas $$\left[\sqrt{x}\right]$$ remains $$1$$ on both sides. Jump of $$+1$$ in the first term is not cancelled → discontinuous.

4. $$x=2$$: $$\dfrac{x^{2}}{2}=2$$ (integer), $$\sqrt{x}\approx1.414\;(non\;integer).$$
Only the first term jumps → discontinuous.

5-8. $$x=\sqrt6,\;\sqrt8,\;\sqrt{10},\;\sqrt{12},\;\sqrt{14}$$ (five points).
In each case $$\dfrac{x^{2}}{2}$$ is an integer, $$\sqrt{x}$$ is not, so only the first term jumps → each is a point of discontinuity.

9. $$x=4$$ (right end-point).  For $$x\to4^{-}$$:
$$\left[\dfrac{x^{2}}{2}\right]=7,\;\left[\sqrt{x}\right]=1\;\Longrightarrow\;f=6.$$
At $$x=4$$: $$\left[\dfrac{x^{2}}{2}\right]=8,\;\left[\sqrt{x}\right]=2\;\Longrightarrow\;f(4)=6.$$
Left-hand limit equals the value, so the function is continuous at $$4$$.

Step 5 : Count the discontinuities.
Discontinuities occur at
$$x=1,\;\sqrt2,\;2,\;\sqrt6,\;\sqrt8,\;\sqrt{10},\;\sqrt{12},\;\sqrt{14}$$ — a total of $$8$$ points.

Therefore, the number of points of discontinuity of $$f(x)$$ on $$[0,4]$$ is $$\mathbf{8}$$.