Let $$A = \{-4, -3, -2, 0, 1, 3, 4\}$$ and $$R = \{(a, b) \in A \times A : b = |a|$$ or $$b^2 = a + 1\}$$ be a relation on $$A$$. Then the minimum number of elements, that must be added to the relation $$R$$ so that it becomes reflexive and symmetric, is _____.

We consider the set $$A = \{-4,-3,-2,0,1,3,4\}$$ and define the relation $$R = \{(a,b) \in A \times A : b = |a|$$ or $$b^2 = a+1\}$$.

To list the elements of $$R$$, we first look at pairs satisfying $$b = |a|$$. For $$a = -4$$ and $$a = -3$$ this gives $$(-4,4)$$ and $$(-3,3)$$. Although $$a=-2$$ yields $$(-2,2)$$, it is not valid since $$2 \notin A$$. The remaining cases $$a=0,1,3,4$$ lead to the pairs $$(0,0),(1,1),(3,3),(4,4)$$, so altogether the condition $$b = |a|$$ contributes the elements $$(-4,4),(-3,3),(0,0),(1,1),(3,3),(4,4)$$ to $$R$$.

Next, we examine the condition $$b^2 = a+1$$, which requires $$a+1$$ to be a perfect square. Checking each element of $$A$$ shows that $$a=0$$ gives $$b^2=1$$ so $$b = \pm 1$$, but since $$-1\notin A$$ only $$(0,1)$$ is valid. For $$a=3$$ one finds $$b^2=4$$ so $$b=\pm2$$, and with $$2\notin A$$ only $$(3,-2)$$ belongs to $$R$$. All other values of $$a$$ in $$A$$ either produce non-integer $$b$$ or lie outside of $$A$$, so they do not contribute new pairs.

Combining both cases yields the relation $$R = \{(0,0),(1,1),(3,3),(4,4),(-4,4),(-3,3),(0,1),(3,-2)\}$$.

To determine how many elements must be added to make $$R$$ reflexive, we note that reflexivity requires $$(a,a)$$ for every $$a\in A$$. The pairs $$(-4,-4),(-3,-3),(-2,-2)$$ are missing from $$R$$, so three elements must be added.

Turning to symmetry, one needs that if $$(a,b)\in R$$ then $$(b,a)\in R$$. From $$(-4,4)$$ we need $$(4,-4)$$; from $$(-3,3)$$ we need $$(3,-3)$$; from $$(0,1)$$ we need $$(1,0)$$; and from $$(3,-2)$$ we need $$(-2,3)$$. None of these are present, so four more elements are required for symmetry.

In total, 3 elements are needed for reflexivity and 4 for symmetry, giving a sum of 7 additional elements.

The answer is $$\boxed{7}$$.