Use mathematical induction to show that (n to the power 5) - n is divisible by 5 for every positive integer n.

Sorry again for the expression. Thanks guys :)

Printable View

- Sep 12th 2006, 11:31 PMsuedenationNumber theory, divisibility, mathematical induction
Use mathematical induction to show that (n to the power 5) - n is divisible by 5 for every positive integer n.

Sorry again for the expression. Thanks guys :) - Sep 13th 2006, 12:29 AMGlaysher
n = 1

1^5 - 1 = 1 - 1 = 0

Assume true for n = k

k^5 - k = 5m for some integer m

Suppose n = k + 1

(k + 1)^5 - (k + 1) = k^5 + 5k^4 + 10k^3 + 10k^2 + 5k + 1 - k - 1

= k^5 - k + 5(k^4 + 2k^3 + 2k^2 + k)

= 5m + 5(k^4 + 2k^3 + 2k^2 + k)

= 5(m + k^4 + 2k^3 + 2k^2 + k)

Hence true for n = k + 1 and by the principle of mathematical induction true for all natural numbers n