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Topic: sum of distinct cubes (Read 763 times) 

Christine
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sum of distinct cubes
« on: Aug 14^{th}, 2014, 9:42am » 
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Cubes with a representation as a sum of distinct cubes : Some end with a 1^3 : 9^3 = 8^3 + 6^3 + 1^3 others don't : 6^3 = 5^3 + 4^3 + 3^3 Is the set of those that don't finite or infinite? How to prove it?


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towr
wu::riddles Moderator Uberpuzzler
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Re: sum of distinct cubes
« Reply #1 on: Aug 14^{th}, 2014, 10:08am » 
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multiply by n^3 if a^3 = b^3 + c^3 + d^3 then (an)^3 = (bn)^3 + (cn)^3 + (dn)^3


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