A prime with 3 factors?

Observe that if $abc$ is prime, then the absolute value of any two of these integers must be equal to $1$ each and the absolute value of the third, a prime. WLOG, we assume $|a| = |b| = 1$ .

Now, if $a=b=1$ , then the common difference is equal to zero and hence we must have $c=1$ as well making $abc = 1$ , which is not a prime.

If $a=b=-1$ , then the common difference is again zero and hence there is no value of $c$ which satisfies $abc=1$ and the arithmetic progression of $a, \ b$ and $c$ simultaneously.

If $a=1$ and $b=-1$ or vice versa, then we have a common difference of $2$ . Thus, in this case, knowing that $a, \ b$ and $c$ are in arithmetic progression with common difference $2$ , we have two values of $c$ which are $-3$ and $3$ . Since $c=3$ gives $abc=-3$ which is not a positive prime number, we check for $c=-3$ . We find that $c=-3$ gives $abc=3$ which is a positive prime. Thus, there exists only one unordered triplet of non-negative integers $a, \ b$ and $c$ which is $\boxed{(1,-1,-3)}$ .