Edwin Clark has pointed out that your 3397 should be 3977, but this is still not prime (= 41 * 97). I believe that your problem(s) have been investigated by people frustrated at being unable to settle the Catalan conjecture, but I'm away from my references at present. Perhaps more later. R. On Sat, 3 May 2003, David Wilson wrote:
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]....
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