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 3^{4}+15^{4} = 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.

I interpret the problem text to mean that if n divides (a^4+b^4) in a nontrivial way then that means n does not divide a^4 or b^4.

Part 3:

If n is twice an odd number, n=2*x, then n divides x^4+x^4 in a nontrivial way.

The remaining even number have n as a multiple of 4, n=4*y. Perfect fourth powers mod 4 are 0 or 1, then if 4*y = a^4+b^4 in a nontrivial way then both a and b are even. If n is 4 then this implies that there is no nontrivial way for 4 to divide a^4+b^4.

What about larger n? Looking at fourth powers mod 3 finds 0 or 1, which again implies a and b are both multiples of 3. This then means that n=12 has no nontrivial answer. Going up through odd values eventually finds one that works, y = 17 = 1^4+2^4. This then can be used to make 68 divide 2^4+4^4, the first nontrivial answer for a multiple of 4.

Part 4:

Every integer n starting with 2 has a nontrivial sum of two fifth powers it divides: 1^5 and (n-1)^5. This follows from the fact that (a+b) divides a*k+b^k whenever k is odd, 5 in this case. Setting a=1 and b=n-1 finishes this answer.