Prime Number

Suppose
A-2
b-5
C-7
And put these no. In equation
Ab+ac=4a^2+8
2×5 +2×7 = 4×2^2 + 8
10+14 = 16+8 24 = 24

Here is the solution.

Taking modulo $a$ , we have that $a$ should be even. Since the only even prime is 2, $a=2$ .

So, substituting in the original equation, we have $b+c=12$ , which implies that $b=7(or)5$ and $a=5(or)7$ .

So, $a*b*c=\boxed{70}$

Let us solve the above equation as a quadratic in $a$ , or $4a^2 - (b+c)a + 8 = 0$ . Solving for $a$ via the Quadratic Formula produces:

$a = \frac{(b+c) \pm \sqrt{(b+c)^2 - 4(4)(8)}}{8} = \frac{(b+c) \pm \sqrt{(b+c)^2 - 128}}{8}$ (i)

and in order for $a$ to be an integer the discriminant must be a perfect square, or:

$(b+c)^2 - 128 = k^2 \Rightarrow (b+c)^2 - k^2 = [(b+c) + k][(b+c) - k] = 2^7$ .

We next check the following four positive divisor pairs of 128:

$(b+c) + k = 128; (b+c) - k = 1 \Rightarrow b+c = \frac{129}{2}$ (NOT VALID);

$(b+c) + k = 64; (b+c) - k = 2 \Rightarrow b+c = 33$ (VALID);

$(b+c) + k = 32; (b+c) - k = 4 \Rightarrow b+c = 18$ (VALID);

$(b+c) + k = 16; (b+c) - k = 8 \Rightarrow b+c = 12$ (VALID).

A final substitution of these values into (i) yields:

$b+c = 33 \Rightarrow a = 8, \frac{1}{4}$ ;

$b+c = 18 \Rightarrow a = 4, \frac{1}{2}$ ;

$b+c = 12 \Rightarrow a = 1, 2$ .

of which only $a = 2, b = 5, c = 7$ is admissible in prime numbers. Hence, $a \cdot b \cdot c = \boxed{70}.$