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.......
the binomial way is simplest,
here's a more convoluted way!
P(x)
$\displaystyle x^5-x$
is divisible by 10
P(x+1)
Does the above statement cause
$\displaystyle (x+1)^5-(x+1)$
to be divisible by 10 ?
Proof
$\displaystyle x^5-x=x(x^4-1)=x(x^2-1)(x^2+1)=x(x+1)(x-1)(x^2+1)$
$\displaystyle (x+1)^5-(x+1)=(x+1)[(x+1)^4-1]=x(x+1)^4-x+(x+1)^4-1$
$\displaystyle =x(x+1)(x+1)(x^2+2x+1)-x+(x+1)(x+1)(x^2+2x+1)-1$
$\displaystyle =x(x+1)(x^3+2x^2+x+x^2+2x+1)-x+(x+1)(x^3+2x^2+x+x^2+2x+1)-1$
$\displaystyle =x(x+1)(x^3+3x^2+3x+1)-x+(x+1)(x^3+3x^2+3x+1)-1$
$\displaystyle =(x+1)\left[x(x^3+3x^2+3x+1)+x^3+3x^2+3x+1-1\right]$
$\displaystyle =(x+1)\left[x^4+3x^3+3x^2+x+x^3+3x^2+3x\right]$
$\displaystyle =(x+1)x(x^3+4x^2+6x+4)$
and since $\displaystyle (x^2+1)(x-1)=x^3-x^2+x-1$
then we have
$\displaystyle (x+1)x(x^3-x^2+x-1+5x^2+5x+5)$
$\displaystyle =(x+1)x\left[(x^2+1)(x-1)+5(x^2+x+1)\right]$
$\displaystyle =(x+1)x(x^2+1)(x-1)+(x+1)(x)5(x^2+x+1)$
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