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Divisibility of N (Posted on 2013-06-15) Difficulty: 3 of 5
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    
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re: No Subject | Comment 2 of 4 |
(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

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