[math-fun] Re: Quadratic question
David Wilson asked:
For which (a, b) do all solutions of a x^2 - b y^2 = [1] have the form (r k, s j) for some fixed r, s > 1?
D. T. Walker, ``On the Diophantine Equation $mX^2 - nY^2 = \pm 1$, American Mathematical Monthly, Volume 74, Issue 5, (May 1967), pp. 504-513, has a theorem that answers Professor Wilson's query. Consider (1) $aX^2 - bY^2 = \pm 1$. Theorem 9 in the above is (a and b are positive integers, not squares (and whose product is not a square)): If (1) is solvable for both of a, b > 1 and has r \sqrt{a} + s \sqrt{b} as its smallest solution; and if i is a nonnegative integer, then the formula r_i \sqrt{a} + s_i \sqrt{b} = (r \sqrt{a} + s \sqrt{b} )^{2i + 1} gives all possible solutions of (1). So r will always divide r_i and s will always divide s_i. Any a, b missing from the list in Professor Clark's reply apparently are not there because either the equation has no solution for those values, or because at least one of r, s is 1. John John Robertson 973-331-9978 jpr2718@aol.com
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]....
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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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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On Sat, 3 May 2003, David Wilson wrote:
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.
David, did this come from the following more general problem? Let E=[e_1,e_2,...,e_k] be a list of integers satisfying 1 <= e_1 <= e_2 <= . . . <= e_k. Let S(E) be the set of all positive integers n such that n = Prod{p_i^e_i} where the p_i's are distinct primes. Say that the integers in S(E) have factorization pattern E. I believe this is a known concept. Which of the sets S(E) contain at least two consecutive integers? Your question is for E = [2,3]. It's easy to find E's which contain 1's for with S(E) contains two consecutive integers, but a search of n <= 2 million shows no such S(E)'s when each e_i > 1. In fact if one checks the minimum gap, n-m, with n,m in S(E), the smallest gap in this range is 5 and it occurs for your case E=[2,3]. Based on this scant evidence maybe [2,3] is the most promising? Is it the case that if 1 is not in E then S(E) contains no consecutive pair of integers? To generalize further: Let S be any set of integers and let f be any integer valued function. Ask for the number of n in S such that f(n) is also in S. The Catalan problem was for S = the set of non-trivial powers and f(n) = n + 1. Maybe this is all old hat? In the above one might replace n+1 by n+2 or n+3 or ...? --Edwin
participants (4)
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David Wilson -
Edwin Clark -
Jpr2718＠aol.com -
Richard Guy