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.
Where m = n
^{4} mod 29, such that 29 is not divisible by n, m is an element of {1,7,16,20,23,24,25}.
As 20+7+1+1 = 29, four 4th powers where modulos are 20, 7, 1 and 1 will be 29. As 25+25+7+1 = 29*2, four 4th powers where the modulos are 25, 25, 7 and 1 will be divisible by 29. Etc.
One can easily find that no combination of any two or three values of m will be a multiple of 29, thus as shown that four values can equal a multiple of 29, the answer to (1.) for 29, is four.
Where m = n^{4} mod 5, such that 5 is not divisible by n, m is an element of {1}. 1+1+1+1+1 = 5. It can be readily seen that to be divisible by 5, there need be at least five 4th powers.
The minimum number required then is that for 29 and four 4th powers.
Edited on June 16, 2013, 6:58 am

Posted by Dej Mar
on 20130616 05:57:08 