in number theory, prove that 5/n^5 - n or in words 5 divides n^5 - n.?

Answer 1

p=n^5 - n
To prove that p/5 is zero or an integer.

Suppose that (n^5 -n) is divisible by 5 . If we can prove that, given (n^5 -n) is divisible by 5 , (n+1)^5 -(n+1) is also divisible by 5 then we can prove the theorem by the method of induction.

Proof:
(n+1)^5 -(n+1)
= n^5+5*n^4
+10*n^3
+10*n^2
+4*n
= (n^5 -n)
+5*n^4
+10*n^3
+10*n^2
+4*n +n
= (n^5 -n)
+ 5* (2*n^3
+ 2*n^2 +n)
Now since (n^5 -n) is divisible by 5 and the rest
5* (2*n^3 + 2*n^2 +n) is obviously divisible by 5 we have proved that,
(n+1)^5 -(n+1) is divisible by 5.

Now put n=1,
1^5 -1 = 0 is divisible by 5
2^5 -2 = 32 -2 = 30 is divisible by 5
Hence, the theorem is proved for n=2,3,4,... for all positive integers.

You can prove it for negative integers too. Use
(n-1)^5 -(n-1)
=n^5
-5*n^4
+10*n^3
-10*n^2
+4*n

Cheers.

Answer 2

Okay. Let n = 5m + k, where m is any integer and k = 0, 1, 2, 3, 4. So, we have:

n^5 - n = m^2(C) + 5m k^4 + k^5 - 5m - k = m^2(C) + 5m(k^4 -1) + k^5 - k.

So we know that if k^5 - k is divisible by 5, n^5 - n is also. Okay, we try all the values for k:

0^5 - 0 = 0
1^5 - 1 = 0
2^5 - 2 = 30
3^5 - 3 = 240
4^5 - 4 = 1020

All are divisible by 5. Therefore QED.

Answer 3

Expressed as a modular equation, this asks us to prove that for all integers n, n^5 = n (mod 5).

The equation holds for 0 and 1, by inspection, so we need only calculate the cases n = 2, 3 and 4 (mod 5), finding that 32 = 2, 243 = 3, and 1024 = 4, all mod 5. QED.

There is an algebraic proof that n^p = n (mod p) for all prime p - look for "Fermat's Little Theorem".

Answer 4

n^5-n = n(n^4-1) = n(n^2-1)(n^2+1)

if n is divisible by 5 then we are done

if n = 5m +/- 1 then n^2 = (25m^2+/-10m + 1) so n^2-1 is divisible by 5

if n = 5m +/-2 then n^2 = (25m^2 +/-20m + 4) so n^2+1 is divisible by 5

Hence it is dvisible by 5 for all cases.
QED

Answer 5

It can also be proved by induction but the details are a bit tedious.