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I am wondering about the integer solutions to ##a^3+b^3+c^3=d^3~##.
By trial and error I stumbled upon ##3^3+4^3+5^3=6^3##. I find this equation remarkable in that not only the four integers are consecutive, but also because the three integers on the left form the well known Pythagorean triplet, ##3^2+4^2=5^2##.
It's like an extension of Fermat's last theorem to what might be called "cubic quartets" by analogy to Pythagorean triplets. If there is one solution, there are probably more. How does one find them other than multiplying an already known solution by a common factor? Has anyone looked into this?
By trial and error I stumbled upon ##3^3+4^3+5^3=6^3##. I find this equation remarkable in that not only the four integers are consecutive, but also because the three integers on the left form the well known Pythagorean triplet, ##3^2+4^2=5^2##.
It's like an extension of Fermat's last theorem to what might be called "cubic quartets" by analogy to Pythagorean triplets. If there is one solution, there are probably more. How does one find them other than multiplying an already known solution by a common factor? Has anyone looked into this?