Prove there are infinitely many positive integers which cannot be represented as a sum of three perfect cubes

hmmm how do you go about this question using contradiction?

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- January 3rd 2010, 09:47 PMusagi_killerSum of perfect cubes
Prove there are infinitely many positive integers which cannot be represented as a sum of three perfect cubes

hmmm how do you go about this question using contradiction? - January 4th 2010, 12:08 AMusagi_killer
Okay, I'm not sure if this is the right way to go about this question, could someone please check it's validity.

I conjecture that all perfect cubes are congruent to either http://stuff.daniel15.com/cgi-bin/ma...pmod%20%7B9%7D

Let http://stuff.daniel15.com/cgi-bin/mathtex.cgi?x be an integer.

Then in http://stuff.daniel15.com/cgi-bin/ma...athbb%7BZ%7D_9 we have:

http://stuff.daniel15.com/cgi-bin/ma...pmod%20%7B9%7D

http://stuff.daniel15.com/cgi-bin/ma...pmod%20%7B9%7D

http://stuff.daniel15.com/cgi-bin/ma...pmod%20%7B9%7D

Now let http://stuff.daniel15.com/cgi-bin/ma...b%5E3,%20c%5E3 be any http://stuff.daniel15.com/cgi-bin/mathtex.cgi?3 perfect cubes.

Thus http://stuff.daniel15.com/cgi-bin/ma...pmod%20%7B9%7D

However the least absolute residue http://stuff.daniel15.com/cgi-bin/mathtex.cgi?%5Cpm%204 is unaccounted for. Since there are infinitely many numbers congruent to http://stuff.daniel15.com/cgi-bin/ma...pmod%20%7B9%7D then there are there are infinitely many positive integers which cannot be represented as a sum of three perfect cubes. - January 5th 2010, 12:45 PMKep
Usagi Killer is correct. Neat method!