Solve the following system of equations:
x + y + z = 6
x2 + y2 + z2 = 18
√x + √y + √z = 4
The third equation limits (x,y,z) to non-negatives at of at most 16. The second limits (x,y,z) to at most sqrt(18).
This limits integer solutions so much that (4,1,1) and permutations are practically trivial.
Call the eqs [1], [2], [3] respectively
([3]^2-[1])/2 yields [4] √xy + √yz + √xz = 5
([1]^2-[2])/2 yields [5] xy + yz + xy = 9
([4]^2-[5])/2 yields x√yz + y√xz + z√xy = 8
√(xyz)(√x + √y + √z) = 8
√(xyz) = 2
[6] xyz = 4
The above solutions fit this, but I'm having trouble showing there aren't others.
If you substitute z=4/(xy) into [1] and [3] you get some nice rings that intersect each other as the above as well.
https://www.desmos.com/calculator/ixoonanijb
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Posted by Jer
on 2024-10-22 14:17:44 |
integer solutions, attempt at reals
