Let the equation $$x^2 + y^2 + px + (1-p)y + 5 = 0$$ represent circles of varying radius $$r \in (0, 5]$$. Then the number of elements in the set $$S = \{q : q = p^2$$ and $$q$$ is an integer$$\}$$ is _________.

We start with the given family of circles

$$x^{2}+y^{2}+px+(1-p)y+5=0.$$

The standard form of the general circle is

$$x^{2}+y^{2}+2gx+2fy+c=0,$$

whose centre is $$(-g,\,-f)$$ and whose radius is $$R=\sqrt{g^{2}+f^{2}-c}\,.$$

Comparing coefficients we obtain

$$2g=p \;\;\Longrightarrow\;\; g=\dfrac{p}{2},$$

$$2f=1-p \;\;\Longrightarrow\;\; f=\dfrac{1-p}{2},$$

and

$$c=5.$$

Substituting $$g,\;f,\;c$$ in the radius formula gives

$$R=\sqrt{\left(\dfrac{p}{2}\right)^{2}+\left(\dfrac{1-p}{2}\right)^{2}-5} =\sqrt{\dfrac{p^{2}}{4}+\dfrac{1-2p+p^{2}}{4}-5}.$$

Simplifying the expression under the square root, we have

$$R=\sqrt{\dfrac{2p^{2}-2p-19}{4}} =\dfrac{1}{2}\sqrt{\,2p^{2}-2p-19\,}.$$

The problem states that the radius varies in the interval $$(0,\,5],$$ therefore

$$0\lt R\le 5 \;\;\Longrightarrow\;\; 0\lt \dfrac{1}{2}\sqrt{\,2p^{2}-2p-19\,}\le 5.$$

Squaring and multiplying by $$4$$ eliminates the square root and the denominator:

$$0\lt 2p^{2}-2p-19\le 100.$$

Thus the two simultaneous inequalities to be satisfied by $$p$$ are

$$\text{(i)}\; 2p^{2}-2p-19 \gt 0,$$

$$\text{(ii)}\; 2p^{2}-2p-119 \le 0.$$

First, solve (i):

$$2p^{2}-2p-19=0 \;\;\Longrightarrow\;\; p=\dfrac{2\pm\sqrt{(-2)^{2}-4\cdot2\cdot(-19)}}{4} =\dfrac{2\pm2\sqrt{39}}{4} =\dfrac{1\pm\sqrt{39}}{2}.$$

Numerically,

$$\dfrac{1-\sqrt{39}}{2}\approx -2.6218,\qquad \dfrac{1+\sqrt{39}}{2}\approx 3.6218.$$

Because the coefficient of $$p^{2}$$ is positive, inequality (i) holds for

$$p\lt \dfrac{1-\sqrt{39}}{2}\qquad\text{or}\qquad p\gt \dfrac{1+\sqrt{39}}{2}.$$

Next, solve (ii):

$$2p^{2}-2p-119=0 \;\;\Longrightarrow\;\; p=\dfrac{2\pm\sqrt{(-2)^{2}-4\cdot2\cdot(-119)}}{4} =\dfrac{2\pm2\sqrt{239}}{4} =\dfrac{1\pm\sqrt{239}}{2}.$$

Numerically,

$$\dfrac{1-\sqrt{239}}{2}\approx -7.2295,\qquad \dfrac{1+\sqrt{239}}{2}\approx 8.2295.$$

Since here the quadratic opens upward and the inequality is “$$\le 0$$”, condition (ii) is satisfied for

$$\dfrac{1-\sqrt{239}}{2}\le p\le\dfrac{1+\sqrt{239}}{2}.$$

The required $$p$$ must satisfy both (i) and (ii) simultaneously. Intersecting the two solution sets gives two disjoint intervals:

$$\Bigl[\dfrac{1-\sqrt{239}}{2},\,\dfrac{1-\sqrt{39}}{2}\Bigr) \;\;\;\text{and}\;\;\; \Bigl(\dfrac{1+\sqrt{39}}{2},\,\dfrac{1+\sqrt{239}}{2}\Bigr].$$

Numerically these are

$$[-7.2295,\,-2.6218)\quad\text{and}\quad(3.6218,\,8.2295].$$

We now form the set

$$S=\{q:q=p^{2} \text{ and } q \text{ is an integer}\}.$$

For any $$q$$ in $$S$$ either $$p=\sqrt{q}$$ or $$p=-\sqrt{q}$$ must lie in at least one of the above intervals.

For $$p\lt 0$$ we require

$$-\sqrt{q}\in[-7.2295,\,-2.6218) \;\;\Longleftrightarrow\;\; \sqrt{q}\in(2.6218,\,7.2295].$$

This implies

$$6.87\lt q\le 52.25,$$

hence all integers

$$q=7,8,9,\ldots,52.$$

For $$p\gt 0$$ we require

$$\sqrt{q}\in(3.6218,\,8.2295],$$

that is

$$13.12\lt q\le 67.77,$$

whence all integers

$$q=14,15,16,\ldots,67.$$

The union of the two integer ranges is simply the continuous block

$$q=7,8,9,\ldots,67.$$

Counting the elements, we get

$$67-7+1=61.$$

So, the answer is $$61$$.