Dr? Roberson: Thank you for your answer. You can just call me Dave Wilson in your posts. I wish I was a professor. All math funners: This question fell out of an investigation as to whether there exist two consecutive integers of the form p^2 q^3 with p, q distinct primes. Let p^2 q^3 + 1 = r^2 s^3. One of p^2 q^3 and r^2 s^3 is even, so one of p, q, r, and s is 2. This leads to four cases [1] 4 q^3 + 1 = r^2 s^3 [2] 8 p^2 + 1 = r^2 s^3 [3] p^2 q^3 + 1 = 4 s^3 [4] p^2 q^3 + 1 = 8 r^2. I was looking at case [4]. This becomes [5] 8 r^2 - q^3 p^2 = 1. Modulo 8, we have q^3 p^2 == -1 ==> q^3 == -1 ==> q == -1. So q is a prime of the form 8k+7, the smallest candidate is q = 7. Putting this in gives [6] 8 r^2 - 343 p^2 = 1. When I solved this computationally, I found the smallest solution was (r, p) = (26041, 3397). Sadly 3397 = 43*79 is not prime. According to John Robertson's reference, all larger solutions will be of the form (r, p) = (26041 j, 3397 k), and so r and p will not be primes. Hence there are no solutions to [6] whith p, q, r distinct primes, and we must look at values of q other than 7. q =23 leads to (r, p) = (39, 1), and 39 is not prime. The next few possibilities for q are 31, 47, 71, 79, 103, 127. In each of these cases, the minimal r and p are at least several tens of digits long, and at least one of them is composite. I do not have the computational machinery to factor these numbers completely, and I do not see any immediate pattern in the known divisors which would allow me to dispense with [4] completely. And then we have [1], [2] and [3]....