The problem Divisibility of 29 essentially asks whether three 4th powers can sum to a multiple of 29 if they are not all multiples of 29.

The entry for 29 in The Penguin Dictionary of Curious and Interesting Numbers by David Wells (1987) contains:

No sum of three 4th powers is divisible by either 5 or 29 unless they all are. [Euler]

1. If three aren't enough, how many 4th powers does it take to be divisible by either 5 or 29 where they aren't all?

2. If possible, find the next number beyond 5 and 29 that does not divide a sum of three 4th powers.

3. Prove every even number takes at most two 4th powers.
For example using 18 we have 3⁴+15⁴ = 50706 = 18*2817

4. What is the largest number of 5th powers whose sum is divisible by a number N where they aren't all divisible by N?

5. Prove that for higher powers, there is no limit to how many numbers it can take.

(In reply to No Subject by Ady TZIDON)

Yes, that's a good formal way to write it.  (Except you squared instead of 4th power.)
Here's another restatement:

For every even number 2k, there exist two numbers a and b not divisible by k, such that a^4+b^4 is divisible by 2k.

In the example 2k=18, a=3, b=6.
3^4+15^4 is divisible by 18.

It may help to look at the problem that inspired this one, which I linked it to.  The parts of this problem were the results of exploring a bit further than the original.

In the original, there are no 3 numbers a, b, c not divisible by 29 for which a^4+b^4+c^4 is divisible by 29. 

My part a asks if 3 numbers a,b,c aren't sufficient then are 4?  or 5? etc.

Posted by Jer on 2013-06-15 19:14:50