The number of distinct real roots of $$x^4 - 4x + 1 = 0$$ is

Given,

$$x^4-4x+1=0$$

Let

$$f(x)=x^4-4x+1$$

Now,

$$f'(x)=4x^3-4=4(x^3-1)$$

Critical point is

$$x=1$$

Also,

$$f''(x)=12x^2>0$$

Hence, $$x=1$$ gives a minimum value.

Now,

$$f(1)=1-4+1=-2<0$$

Also,

$$\lim_{x\to\pm\infty}f(x)=+\infty$$

Further,

$$f(0)=1>0,\qquad f(2)=16-8+1=9>0$$

So, one root lies in $$(0,1)$$ and another root lies in $$(1,2)$$.

Also,

$$f(-1)=1+4+1=6>0$$

and the function is decreasing for $$x<1$$.

Hence, there is no negative root.

Therefore, the equation has exactly $$\boxed{2}$$ distinct real roots.