The question is whether x^4+y^4+z^4 could be a multiple of 29 but not all of x,y,z multiples of 29.
Consider the residuals (mod 29) of n^4 for n=0,1,2,...,14
0,1,16,23,24,26,20,23,7,7,24,25,1,25,20
The surprising thing about this set is that no three of them sum to 0 (mod 29) except for 0+0+0
In other words, the solutions given in my other post are all of the solutions. The determination is, therefore:
YES. If 29 divides the sum, then 29^4 will always divide the sum.
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Posted by Jer
on 2013-05-15 12:40:39 |
Rest of solution
