This is the solution to the problem of Week #1 due to Robert Dawson: <BR><BR>

a,b,c are in arithmetic progression. n is the common difference. <BR><BR>

Claim a=3.
All or none of {a,b,c}  are congruent mod 3.  If they are all ==0, they are
3, 3, and 3; but that makes d = 3*3-3 be 6, which is composite.
Suppose all == 1 or 2. Then n==0,   ac==d, d==1. But d-a-b-c is congruent to
1, so its root is congruent to 1 or 2 =><=
Conclude exactly  one of a,b,c == 3.   To be prime, it must be a=3. <BR><BR>

So we have  a=3;  b=3+n;  c=3+2n. <BR><BR>

We also have ac-d = 9+6n-d = n, d = 5n+9. <BR><BR>

We also have n^2 = d-a-b-c = 5n-9 -9 -3n = 2n. n = 0 or 2  and we have ruled
out n=0 so n=2. Our primes are 3,5,7, and 3*7-2 = 19. <BR><BR>

Now, the product of these is 1995 (a year when it's plausible that you were
in university).  As problem setters frequently factorize the year number in
the hope of inspiration, the passage from 1995 to {3,5,7,19} and to the
arithmetic sequence is plausible. <BR><BR>

With n=4 we get {3,7,11,29} which have b-a = c-b = ac-d. <BR>
With n=8 we get {3,11,19,97} which have b-a = c-b, d=a+b+c+n^2 <BR><BR>

<i> Mitja remarks here that b-a=c-b=ac-d does not imply that a=3. </i>  For example
(a,b,c,d)= (5, 11, 17, 79) also works. <BR><BR>

<i> The original posting contained a few typos.  They were noticed by 
Owen Sharpe.  Thank you! </i>