We can deal with [2] 8p^2 + 1 = r^2 s^3 Either p = 3, which is easily dismissed, or p^2 = 1 mod 3 and r or s = 3. 8p^2 - 27r^2 = -1 fails mod 8, and the elliptic curve 8p^2 = 9s^3 - 1 has the solution p = s = 1 and the nearest approach to another integer one is p = 11/2, s = 3. It does have an infinity of rational solutions. Will similar considerations deal with [1] and [3] ? 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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