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Sum of Cubes (Posted on 2013-10-23) Difficulty: 2 of 5
Express each integer between 1 and 1000 as a sum of four cubes of integers (which can be positive, negative, or zero).

No Solution Yet Submitted by Danish Ahmed Khan    
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No Subject | Comment 4 of 5 |
(In reply to Computer findings, leaving 45 cases for analytic methods by Charlie)

31 = (-56)^3 + (-41)^3 + (-26)^3 + 64^3 
148 = (-104)^3 + 91^3 + 73^3 + (-26)^3 
265 = (-26)^3 + 7^3 + (-53)^3 + 55^3
284 = (-130)^3 + 5^3 + 80^3 + 119^3
356 = (-112)^3 + (-103)^3 + (-85)^3 + 146^3 
382 = (-149)^3 + 34^3 + 91^3 + 136^3

571 = (-170)^3 + (-260)^3 + (-56)^3 + 283^3

580 = (-77)^3 + 73^3 + 16^3 + 40^3 
598 = (-131)^3 + (-47)^3 + (-5)^3 + 133^3 
626 = (-142)^3 + 131^3 + 80^3 + 47^3 
724 = (-134)^3 + 97^3 + (-53)^3 + 118^3
 
725 = (-67)^3 + 59^3 + 56^3 + (-43)^3 
787 = (-131)^3 + (-47)^3 + 4^3 + 133^3 
949 = (-101)^3 + 109^3 + 37^3 + (-68)^3

It appears that all integers can be expressed as a sum of four integer cubes.
WHY?


  Posted by Benny on 2013-10-25 14:52:35
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