I agree.

A slightly different approach:

If a number is of the form 4k^6, then it is worth {0,4,36}, mod64.

        
If a number is of the form 2k^6, then it is worth either 0, or  (16n-14), mod64.   

If a number is of the form k^6, then it is worth either 0, or  (8n-7), mod64. 

       
But there is no combination of numbers of the form (16n-14) and (8n-7) which gives a total of {0,4,36}. Hence each of a,b,c, must be worth 0, mod64, and so even, say (2x),(2y),(2z).

(2x)^6+2(2y)^6=4(2z)^6 resolves to x^6+2y^6=4z^6, and so on ad infinitum, with no solution other than a=b=c=0.

Edited on September 30, 2013, 3:11 am

Posted by broll on 2013-09-02 05:02:05