A prime number p is called special if there e | vedclass

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(B) Given $p = p_1 + p_2 = p_3 - p_4$ where $p, p_1, p_2, p_3, p_4$ are primes.Case $I$: If all $p_1, p_2, p_3, p_4$ are odd,then $p_1 + p_2$ is even,

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(B) Given $p = p_1 + p_2 = p_3 - p_4$ where $p, p_1, p_2, p_3, p_4$ are primes.Case $I$: If all $p_1, p_2, p_3, p_4$ are odd,then $p_1 + p_2$ is even,which implies $p$ is even. Since $p$ is a prime,$p = 2$. However,$p_1 + p_2 = 2$ is impossible for primes $p_1, p_2 \ge 2$.Case $II$: At least one of $p_1, p_2$ must be $2$. Let $p_2 = 2$. Then $p = p_1 + 2$. Since $p$ and $p_1$ are primes,$p_1$ must be odd (unless $p_1=2$,then $p=4$ not prime). If $p_1$ is odd,$p$ is odd.From $p = p_3 - p_4$,we have $p_3 = p + p_4$. Since $p$ is odd,for $p_3$ to be prime,$p_4$ must be $2$ (if $p_4$ were odd,$p_3$ would be even and $>2$,hence not prime).Thus,$p = p_1 + 2$ and $p = p_3 - 2$,which means $p_3 = p + 2$ and $p_1 = p - 2$.We need $p-2, p, p+2$ to be primes. This is a prime triplet of the form $(n, n+2, n+4)$. The only such triplet is $(3, 5, 7)$.Therefore,$p = 5$ is the only solution.The number of special primes is $1$.

List-$I$	List-$II$
$(i)$ $\alpha = \beta$	$(A)$ $(ac^2)^{1/3} + (a^2c)^{1/3} + b = 0$
$(ii)$ $\alpha = 2\beta$	$(B)$ $2b^2 = 9ac$
$(iii)$ $\alpha = 3\beta$	$(C)$ $b^2 = 6ac$
$(iv)$ $\alpha = \beta^2$	$(D)$ $3b^2 = 16ac$
	$(E)$ $b^2 = 4ac$
	$(F)$ $(ac^2)^{1/3} + (a^2c)^{1/3} = b$

$A$ prime number $p$ is called special if there exist primes $p_1, p_2, p_3, p_4$ such that $p = p_1 + p_2 = p_3 - p_4$. The number of special primes is

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