Key idea. Express the square of the prime as a product using the given equation.

1. Write \(p=a-b\), which is a positive prime, and \(s=a+b\). Then

   \[p^2=(a-b)^2=s^2-4ab.\]

2. The hypothesis gives \(ab=3c^2-cs\). Substitute and factor:

   \[p^2=s^2+4cs-12c^2=(s-2c)(s+6c).\]

3. The second factor is positive. Since their product is positive, the first is positive too. They are distinct, because their difference is \(8c>0\).

4. The positive factor pairs of a prime square \(p^2\), in increasing order, are \((1,p^2)\) and \((p,p)\). Our factors are distinct, so

   \[s-2c=1,\qquad s+6c=p^2.\]

5. Subtract these two equations: \(8c=p^2-1\). Hence \(8c+1=p^2\), the square of the prime \(a-b\).

Remember. A prime square has very few factor pairs. Positivity and the order of the factors decide which pair is possible.