Fred Helenius found (15+14i)^5 + (5-18i)^5 = (18-7i)^5 + (2+3i)^5 --- Joerg Arndt <jj@suse.de> wrote:
p^5 + q^5 == r^5 + s^5 == v
[p, q, r, s] v
[1 + 3/2*I, 1 - 3/2*I, 3/2, 1/2] 61/8 [1 + 7/5*I, 1 - 7/5*I, 1 + 1/5*I, 1 - 1/5*I] 152/125 [1 + 9/4*I, 1 - 9/4*I, 11/4, -3/4] 20101/128 [1 + 27/10*I, 1 - 27/10*I, 33/10, -13/10] 387641/1000 [1 + 51/10*I, 1 - 51/10*I, 59/10, -39/10] 6247001/1000 [1 + 17/13*I, 1 - 17/13*I, 1 + 7/13*I, 1 - 7/13*I] -84488/28561
Note that the solutions are semi-trivial: the sum of the lhs. terms is equal to those on the rhs.
My parametric gives square roots in general. Need more time to check whether I can repair that. (My documentation and code about the stuff is extremely messy).
I am pretty sure there are solutions for higher powers, too. If there is interest, I'll do a search when time permits.
* Joerg Arndt <jj@suse.de> [Jan 09. 2003 13:09]:
as to: <CITE http://www.mathpuzzle.com/Gaussians.html> (3+13i)^3 + (7+i)^3 = (3+10i)^3 + (1+10i)^3 and (6+3i)^4 + (2+6i)^4 = (4+2i)^4 + (2+i)^4 Is there something similar for Gaussian fifth powers? See http://euler.free.fr/details.htm for details on this problem in the integers. I offer $20 for a nontrivial solution with proper Gaussian integers </CITE http://www.mathpuzzle.com/Gaussians.html>
I have a parametric solution for the fifth power (at home). I'll dig it up and send the solution to the list.
all the best, jj
* ed pegg <ed@mathpuzzle.com> [Jan 06. 2003 20:26]:
There are various number theoretic problems in the Gaussian Integers that I haven't been able to solve.
http://www.mathpuzzle.com/Gaussians.html
If any of these matters are resolved, or if you have solutions to any of them, please let me know.
--Ed Pegg Jr, www.mathpuzzle.com
-- p=2^q-1 prime <== q>2, cosh(2^(q-2)*log(2+sqrt(3)))%p=0 Life is hard and then you die.
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-- p=2^q-1 prime <== q>2, cosh(2^(q-2)*log(2+sqrt(3)))%p=0 Life is hard and then you die.
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