Answer. No. Solution. Obviously, after the first step the sum of $ a,b,c$ and $ d$ is zero.So after $ 1996$ steps we will have that $ a+ b + c + d = 0$. So we get the following relations: $ ab - cd = ab + c(a + b + c) = (c + a)(c + b)$ $ ac - bd = ac + b(a + b + c) = (b + a)(b + c)$ $ bc - ad = bc + a(a + b + c) = - (a + b)(a + c)$ So we get that: $ |ab - cd|\cdot|ac - bd|\cdot|ad - bc| = (a + b)^2(b + c)^2(c + a)^2$ But the product of three primes can't be a perfect square, so the answer is no.

Opinion about the following problem

Indeed, the problem is quite simple. If I am not missing something, it's clear that after at most four moves all the numbers must be even (simple casework), and the numbers will thereafter remain even; then each of the differences in question is divisible by 4, so they are obviously not prime. QED

Really.....when you see this kind of problems you feel that the quality of ISL is decreasing,but when you see certain other problems you really want to have a go at the ISL problems

Trivial. Suppose that at the $1995^{\text{th}}$ step the numbers on the circle are $w,x,y,z$ (in that order). Then we have $$|bc-ad|=|(x-y)(y-z)-(w-x)(z-w)|=|(w-y)(w+y-x-z)|$$$$|ac-bd|=|(w-x)(y-z)-(x-y)(z-w)|=|(w-y)(x-z)|$$$$|ab-cd|=|(w-x)(x-y)-(y-z)(z-w)|=|(x-z)(w+y-x-z)|$$For these three numbers to be prime, we must have either $|w+y-x-z|=|w-y|=1$ or $|w+y-x-z|=|x-z|=1$ or $|x-z|=|w-y|=1$, all of which contradict the fact that $1$ is not a prime. Done.