Re: [math-fun] Is this possible?
Fred wrote: << A simple-minded search for solutions modulo primes of form p = 12 k + 1 yields 2640 solutions when p = 157 , whence the density of solutions cannot exceed 0.00000435 . As the number of solutions looks to be something of order very roughly p ^ (3/2) , the density is very likely actually zero. I haven't checked whether any of these Z_p solutions is actually a solution over Z. On 6/5/10, Hans Havermann <pxp@rogers.com> wrote: << Quadruples {a,b,c,d}, a>b>c>d>0, where a+b, a+c, a+d, b+c, b+d, c+d are all squares, a+b+c, a+b+d, a+c+d, b+c+d are all cubes, and a+b+c+d is a fourth power.
Very interesting! I was thinking the number of conditions (11) is a lot bigger than the number of unknowns (4), so was guessing there might be no solution. Clearly if modulo any number at all there are no solutions, then there are none over Z, so I was wondering about p = 7, which looked like the first plausible case where no solutions might exist. So, Fred, can you run your program on primes of other forms than 12k + 1, like 8k + 7, or just all primes up to some size ? Or are primes 12k + 1 easier to solve Hans's equations for ? --Dan ________________________________________________________________________________________ "Outside of a dog, a book is man's best friend. Inside of a dog, it's too dark to read." --Groucho Marx
I picked on p = 1 (mod 12) simply so that cubic and quartic residues would be sparser, rejecting a greater proportion of cases. The algorithm is trivial: initially compute the sets of quadratic, cubic and quartic residues mod p; for each decreasing [a,b,c,d], test whether every sum mod p lies in the appropriate set. For p = 3 find 3 solutions --- [0,0,0,0], [1,0,0,0], [2,2,2,1]; for p = 5 find 7; for p = 7 find 3 --- [0,0,0,0], [1,0,0,0], [2,2,2,2]; for p = 11 find 41; for p = 13 find 4 --- [0,0,0,0], [1,0,0,0], [12,1,0,0], [8,8,5,5]; for p = 17 find 111; for p = 19 find 7; for p = 23 find 331; for p = 37 find 36; for p = 61 find 141; for p = 73 find 380; for p = 157 find 2640; [The zero solution is irrelevant of course, since a common factor can be divided out.] WFL On 6/6/10, Dan Asimov <dasimov@earthlink.net> wrote:
Fred wrote:
<< A simple-minded search for solutions modulo primes of form p = 12 k + 1 yields 2640 solutions when p = 157 , whence the density of solutions cannot exceed 0.00000435 . As the number of solutions looks to be something of order very roughly p ^ (3/2) , the density is very likely actually zero.
I haven't checked whether any of these Z_p solutions is actually a solution over Z.
On 6/5/10, Hans Havermann <pxp@rogers.com> wrote: << Quadruples {a,b,c,d}, a>b>c>d>0, where a+b, a+c, a+d, b+c, b+d, c+d are all squares, a+b+c, a+b+d, a+c+d, b+c+d are all cubes, and a+b+c+d is a fourth power.
Very interesting!
I was thinking the number of conditions (11) is a lot bigger than the number of unknowns (4), so was guessing there might be no solution. Clearly if modulo any number at all there are no solutions, then there are none over Z, so I was wondering about p = 7, which looked like the first plausible case where no solutions might exist.
So, Fred, can you run your program on primes of other forms than 12k + 1, like 8k + 7, or just all primes up to some size ? Or are primes 12k + 1 easier to solve Hans's equations for ?
--Dan
________________________________________________________________________________________ "Outside of a dog, a book is man's best friend. Inside of a dog, it's too dark to read." --Groucho Marx
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