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Can you solve the equation x³ + y³+ z³= k, where x, y, z are integers and k is an integer from 1 to 100?
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16 Answers

Ronald Deep, worked at The University of Dayton
Answered September 7, 2019 · Upvoted by Allan Gilbertson, M.A. Applied Mathematics, University of Maryland, College Park
[(k,x,y,z) for k in range(1,100) for x in range(10) for y in range(10) for z in range(10) if x**3 + y**3 + z**3 == k]
[(1, 0, 0, 1), (1, 0, 1, 0), (1, 1, 0, 0), (2, 0, 1, 1), (2, 1, 0, 1), (2, 1, 1, 0), (3, 1, 1, 1), (8, 0, 0, 2), (8, 0, 2, 0), (8, 2, 0, 0), (9, 0, 1, 2), (9, 0, 2, 1), (9, 1, 0, 2), (9, 1, 2, 0), (9, 2, 0, 1), (9, 2, 1, 0), (10, 1, 1, 2), (10, 1, 2, 1), (10, 2, 1, 1), (16, 0, 2, 2), (16, 2, 0, 2), (16, 2, 2, 0), (17, 1, 2, 2), (17, 2, 1, 2), (17, 2, 2, 1), (24, 2, 2, 2), (27, 0, 0, 3), (27, 0, 3, 0), (27, 3, 0, 0), (28, 0, 1, 3), (28, 0, 3, 1), (28, 1, 0, 3), (28, 1, 3, 0), (28
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Imad M. I. Zoughaib, 10 years in study and researches
Answered December 27, 2019
We understand x , y and z as integers belonging to IZ => negative and positive Natural integers.
x3+y3+z3=kx3+y3+z3=k with k is integer from 1 to 100
solution x=0 , y=0 and z=1 and k= 1
For K= 1 , we have the following solutions (x,y,x) = (1,0,0) ; or (0,1,0) ; or (0,0,1) ,
For k =1 also (9,-8,-6) or (9,-6,-8) or (-8,-6,9) or (-8,9,-6) or (-6,-8,9) or (-6,9,8)
And (-1,1,1) or (1,-1,1)
=>(x+y)3−3x2−3xy2+z3=k=>(x+y)3−3x2−3xy2+z3=k
=>(x+y+z)3−3(x+y)2.z
