Dividable by 5

and fermat shows that any time you choose the power to be a prime, it works

and this is what leads mathematicians into obsessions with primes

Originally posted by joe shmo
and this is what leads mathematicians into obsessions with primes

indeed 🙂 also because they can be used to encrypt high security information... lots of money in new primes hehe

I vote for Tricky Dicky's solution. Simple, complete and using results that other mathematicians have found in the past 🙂

Originally posted by coquette
I'm so not a mathematician but I think that this problem is simple and obvious. the number raised to any power with the number subtracted is obviously still a multiple of the number ....

In maths, you can't just say something is true becuase it looks true. For example:

31 is prime
331 is prime
3 331 is prime
33 331 is prime
333 331 is prime
3 333 331 is prime
33 333 331 is prime
333 333 331 is NOT prime = 17 * 19 607 843

Well, I think that it's quite easy to prove. If all n^5 - n where n is from 1 to 9 are divisible with 5, then every number that has 1 to 9 as the last digit is in this form divisible with 5. (with 0 it's REALLY obvious, the number will always end with 0, hence, will be divisible with 0.)
1^5 - 1 = 0 => I'm not the one to say whether this is divisible with 5 or not but if it was 21, then the result would end with zero, hence it would be divisible. So this case (n=1) should be considered an exception.
2^5 - 2 = 30 => divisible with 5
3^5 - 3 = 270 => divisible with 5
4^5 - 4 = 1020 => divisible with 5
5^5 - 5 = 3120 => divisible with 5
6^5 - 6 = 7770 => divisible with 5
7^5 - 7 = 16800 => divisible with 5
8^5 - 8 = 32760 => divisible with 5
9^5 - 9 = 59040 => divisible with 5

So there you have it. Any number that end's with a digit from this list is in form n^5 - n divisible with 5. The exceptional cases are n=0 and n=1. I'm not sure though that this can be proven algebraically.

EDIT: My bad. That doesn't work.

Originally posted by Agerg
Proof by induction that for all n >= 0, 5|(n^5 - n)

Base case
let n = 0
0^5 - 0 = 0 is divisible by 5 so result holds

Assume by inductive hypothesis result holds for n = k, ie; 5|(k^5-k)

Let n = k+1

(K+1)^5 - (k+1)
=k^5 +5k^4 +10k^3 + 10k^2 + 5k +1 - (k+1)
=k^5 +5(k^4 +2k^3 +2k^2 + k) - k
=(k^5 - k) + 5(k^4 +2k^3 +2k^2 + k) is divisible by 5 ...[text shortened]... ion on n it is true that for all n >= 0, 5|(n^5 - n)

a similar argument can be used for n < 0

This argument already nails it, and not just for 5 either. It proves that (n^5 - n) is divisible by 10 as well, by the same reasoning.

For n=0, 0^5 - 0 = 0 is divisible by 10.

Now if n holds for some k (which it does), does it hold for k+1?
Again, (k+1)^5 - (k+1) = (k^5 - k) + 5(k^4 +2k^3 +2k^2 + k)

We know the first part is divisible by 10, but we can also prove that a 2 can come out of (k^4 +2k^3 +2k^2 + k), thus making the whole expression divisible by 10.

2k^3 and 2k^2 are obviously divisible by 2. Now, if k is even, then both k and k^4 are divisible by 2, so it works. If k is odd, then neither k nor k^4 is divisible by 2, which means their sum is. Therefore, for any n, n^5 - n is divisible by 10.

the proof for all inices being prime

k=1
k^p-k = 1^p-1 = 1-1 = 0 : 0 is divisible by p

(k+1)^p -(k+1) = sum_1 -(k+1)
sum_1= sum[c_j * k^(p-j) | j = (0,1,2,...,p-1,p) : c_j = {p! / ( (p-j)! * j! )} ]
c_j = [ 1 , p , p*(p-1)/2 , p*(p-1)(p-2)/2*3 , ....]
if and only if p is mutually prime with all integers less than p then c_j is divisible by p for all j not 0 or p
therefore : c_j is divisible by p for all p that are prime
sum_1 = k^p +1 + p*sum_2
sum_2 = sum[c_j * k^(p-j) | j = (1,2,...,p-1) : c_j = {(p-1)! / ( (p-j)! * j! )} ]
(k+1)^p -(k+1) =
k^p +1 + p*sum_2 -(k+1) =
k^p -k + p*sum_2 = Q
since [k^p -k] and [p*sum_2] are both divisible by p then Q is divisible by p

therefore :
k = integer
p = prime

then k^p-k is divisible by p