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 Divisibility of N (Posted on 2013-06-15)
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.

 No Solution Yet Submitted by Jer No Rating

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 Parts 3 and 4 Comment 4 of 4 |
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.

 Posted by Brian Smith on 2017-06-26 23:40:11

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