Thread: Mathematical Induction Problem!

I gotta prove that X^5 - X is a multiple of 10 for all natural numbers through mathematical induction... Plz Help!

Originally Posted by chlwlsdud

I gotta prove that X^5 - X is a multiple of 10 for all natural numbers through mathematical induction... Plz Help!

Show that it is true for x=1 (or 2 even)
Assume true for x=k ie assume k^5-k=10c (where c is an integer)
Then look at
(k+1)^5 - (k+1)
Expand out the binomial, the ones will cancel.
Replace the terms k^5-k with 10c.
At this stage you shpuld have:
10c + 5k^4 +10k^3 +10k^2 +5k
=10(c + k^3 + k^2) + 5 (k^4 + k)

The first part of this is obviously a multiple of 10.
The last bit is obviously a multiple of 5 but also k^4+k is always even
(if k is odd, k^4 is odd and so k^4+k is even; if k is even, k^4 is even and so k^4+k is even), so the last bit is a multiple of 5 and 2 ie a multiple of 10.......

Originally Posted by chlwlsdud

I gotta prove that X^5 - X is a multiple of 10 for all natural numbers through mathematical induction... Plz Help!

the binomial way is simplest,
here's a more convoluted way!

P(x)



is divisible by 10

P(x+1)

Does the above statement cause



to be divisible by 10 ?

Proof

















and since

then we have







The first part of the sum is divisible by 10 if the hypothesis is true.
The second part is a multiple of 5, but is also a multiple of 10 since if x is odd, x+1 is even and vice versa.

Test for n=0, 1, 2

0-0=0 divisible by 10
1-1=0 divisible by 10
32-2=30 divisible by 10.

Hence the hypothesis is true