Obvious solutions
x=y=z=0
If x=y=z=29 then the sum is 3*29^4 which is divisible by 29^4.
You can generalize this result to:
(29a)^4 + (29b)^4 + (29c)^4 = 29^4(a^4+b^4+c^4) for any integers a,b,c.
So the answer to the problem is either "sometimes" or "yes, always"
Are there solutions where x,y,z are not all multiples of 29?

Posted by Jer
on 20130515 12:25:00 