[math-fun] Is this possible?
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.
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. Fred Lunnon 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.
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It seems hard enough to satisfy just a small subset of the conditions. Ignoring the conditions about squares, I found: (a,b,c,d) = (1468664, 811728, 705592, 125712) a+b+c = 144^3, a+b+d = 134^3, a+c+d = 132^3, b+c+d = 118^3 a+b+c+d = 42^4 I would be really surprised if solutions satisfying all conditions exist, but I don't see an obvious obstruction. Jim On Sat, Jun 5, 2010 at 10:49 AM, 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.
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participants (3)
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Fred lunnon -
Hans Havermann -
James Buddenhagen