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?
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Posted by Jer
on 2013-05-15 12:25:00 |