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
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.
Posted by Jer
on 2013-05-15 12:40:39