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Math Help - Mathematical Induction Problem!

  1. #1
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    Mathematical Induction Problem!

    I gotta prove that X^5 - X is a multiple of 10 for all natural numbers through mathematical induction... Plz Help!
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  2. #2
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    Quote Originally Posted by chlwlsdud View Post
    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.......
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  3. #3
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    Quote Originally Posted by chlwlsdud View Post
    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)

    x^5-x

    is divisible by 10

    P(x+1)

    Does the above statement cause

    (x+1)^5-(x+1)

    to be divisible by 10 ?

    Proof

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

    (x+1)^5-(x+1)=(x+1)[(x+1)^4-1]=x(x+1)^4-x+(x+1)^4-1

    =x(x+1)(x+1)(x^2+2x+1)-x+(x+1)(x+1)(x^2+2x+1)-1

    =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

    =x(x+1)(x^3+3x^2+3x+1)-x+(x+1)(x^3+3x^2+3x+1)-1

    =(x+1)\left[x(x^3+3x^2+3x+1)+x^3+3x^2+3x+1-1\right]

    =(x+1)\left[x^4+3x^3+3x^2+x+x^3+3x^2+3x\right]

    =(x+1)x(x^3+4x^2+6x+4)

    and since (x^2+1)(x-1)=x^3-x^2+x-1

    then we have

    (x+1)x(x^3-x^2+x-1+5x^2+5x+5)

    =(x+1)x\left[(x^2+1)(x-1)+5(x^2+x+1)\right]

    =(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
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