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   Author  Topic: sum of distinct cubes  (Read 780 times)
Christine
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sum of distinct cubes  
« on: Aug 14th, 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
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Re: sum of distinct cubes  
« Reply #1 on: Aug 14th, 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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Wikipedia, Google, Mathworld, Integer sequence DB
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