prove that by using mathematical induction(5 divides n the power 5 minus n whenever n is a non negative integer
Base case: When n=0, we have 5^0 - 0 = 1 - 0 = 1, which is divisible by 5.
Inductive step: Assume that for some non-negative integer k, 5 divides k^5 - k. We want to show that 5 divides (k+1)^5 - (k+1).
Expanding (k+1)^5, we get:
(k+1)^5 = k^5 + 5k^4 + 10k^3 + 10k^2 + 5k + 1
Subtracting k+1 from this expression, we get:
(k+1)^5 - (k+1) = k^5 + 5k^4 + 10k^3 + 10k^2 + 4k
We can factor out a k from this expression:
(k+1)^5 - (k+1) = k(k^4 + 5k^3 + 10k^2 + 10k + 4)
Since k^5 - k is divisible by 5 by our assumption, we know that k^5 - k = 5m for some integer m. We can use this fact to simplify our expression:
(k+1)^5 - (k+1) = k(k^4 + 5k^3 + 10k^2 + 10k + 4) = k(5m + 4)
Since 5 divides 5m and 5 divides 4 (since 4 = 5 - 1), we know that 5 divides 5m + 4, which means that 5 divides (k+1)^5 - (k+1).
Therefore, by mathematical induction, we have proven that 5 divides n^5 - n for any non-negative integer n.
To prove this statement using mathematical induction, we need to show that it is true for a base case (n=0), and if the statement is true for an arbitrary non-negative integer k, then it is true for k+1.
Base case (n=0):
0^5 - 0 = 0, which is divisible by 5.
Inductive step:
Let's assume the statement is true for some non-negative integer k, i.e., 5 divides k^5 - k. We need to prove that it is true for k+1, i.e., 5 divides (k+1)^5 - (k+1).
(k+1)^5 - (k+1) = k^5 + 5k^4 + 10k^3 + 10k^2 + 5k + 1 - k - 1
= k^5 - k + 5k^4 + 10k^3 + 10k^2 + 5k
= (k^5 - k) + 5(k^4 + 2k^3 + 2k^2 + k)
By our induction assumption, 5 divides k^5 - k. Also, 5 clearly divides the second term 5(k^4 + 2k^3 + 2k^2 + k), since this term has a factor of 5. Since 5 divides both terms, it must also divide their sum:
5 divides [(k^5 - k) + 5(k^4 + 2k^3 + 2k^2 + k)] = (k+1)^5 - (k+1)
So, if 5 divides k^5 - k, then 5 divides (k+1)^5 - (k+1). By mathematical induction, this statement is true for all non-negative integers n.