(x+y)^2=x^3+y^3
The above is a d2 or d3 problem if one tries simulation or algebraic solution.
Now solve it d1!
Non negative integers, of course !
Source: appeared as a question and an answer(super complicated!)in
QUORA
The LHS is a perfect square and the RHS is the sum of two cubes. There aren't many options:
0^2=0^3+0^3 --> (0,0)
1^2=0^3+1^3 --> (1,0), (0,1)
2^2=none
3^2=1^3+2^3 --> (1,2), (2,1)
4^2=2^3+2^3 --> (2,2)
There are larger squares that are the sum of two cubes but they won't be solutions because the right side will require larger numbers than the left.
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Higher D addendum:
8^2=4^3+0^3 but 8 is not 4+0
24^2 =8^3+4^3 but 24 is not 8+4
https://oeis.org/A217248
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Posted by Jer
on 2021-12-31 09:19:03 |
Another D1 approach
