Consider the equation x^3 + y^3 + z^3 = k, where x, y, z, and k are positive integers. Find the smallest positive integer value of k for which there are no positive integer solutions (x, y, z) that satisfy the equation.
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satisfy me by your answer and Get a Brainliest
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Verified answer
Answer:
33=8 866 128 975 287 5283+(−8 778 405 442 862 239)3+(−2 736 111 468 807 040)3
42=(−80 538 738 812 075 974)3+80 435 758 145 817 5153+12 602 123 297 335 6313
Answer:
Presumably the OP is asking how to solve x3+y3+z3=k
in integers.
The answer is of course to ask Andrew Booker of the University of Bristol. Professor Booker recently found the answer for k=33
and (together with Andrew Sutherland) k=42,
which were the only remaining unknown ones under 100. The pair also found a third way to make k=3
(which was a famous unsolved problem) and a few more in the 100–1000 range.
33=8 866 128 975 287 5283+(−8 778 405 442 862 239)3+(−2 736 111 468 807 040)3
42=(−80 538 738 812 075 974)3+80 435 758 145 817 5153+12 602 123 297 335 6313
The short answer of how to find these is to start from a good grasp of number theory to come up with an efficient search computation and then apply a few million hours of compute time. Booker and Sutherland used the Charity Engine, which is a huge network of home computers available for compute jobs. It’s generally rented by universities and corporations and the proceeds go to charity.
It’s worth pointing out that for k≡±4(mod9)
there are no integer solutions. That’s pretty easy to show; there are only 9 numbers mod 9, which are 0,±1,±2,±3,±4
whose cubes mod 9 are 0,±1,∓1,0,±1
respectively. Since each cube can only be -1, 0 or 1 the sum of three of them is in the range -3 to 3 so can’t be 4 or -4.