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X^5 cong. x(mod10) need proof

  1. Sep 29, 2009 #1
    1. The problem statement, all variables and given/known data
    Number theory problem. we are just doing modular division and congruence theory.
    x^5==x(mod10) ==> 10|x^5-x


    2. Relevant equations
    induction. let x=1,x=x+1


    3. The attempt at a solution
    let x=1 1^5==1(mod10) this is trivial but 2^5==2(mod10)...
    let x=x+1
    (x+1)^5==(x+1)mod10 ==> 10|(x+1)^5-(x+1)
    (x+1)^5=x^5+(....)+1 ==> 10|x^5-x+(...)

    now we know that 10|x^5-x by assumption. so that part is done just need to show that 10|(...)

    (...)=5x^4+10x^3+10x^2+5x

    obviously 10|10x^3+10x^2 so we require 10|5x^4+5x.

    10|5x(x^3+1). if x is even than we can factor out a 2 and get 10|10x/2(x^3+1) and be done.
    if x is odd. then x^3+1 is always even and we can factor out a 2 and get 10|10x(x^3+1)/2.

    does this last part need proving? can you give me a better way of proving this? induction often works, but i like finding clever ways of doing proofs instead of brute force with induction.
     
  2. jcsd
  3. Sep 29, 2009 #2

    VietDao29

    User Avatar
    Homework Helper

    Your idea is correct, however, your presentation is a real mess!!!!

    There are a few things you need to remember is:

    • Do not start from what you need to prove. Instead, start from what you have, and from there deduce to what you need to prove.
    • Unless you are positively sure about what [itex]\Rightarrow[/itex], [itex]\Leftrightarrow[/itex] mean; please use your words instead.

    -----------------------

    What you should have presented is:

    1. For n = 1, true.
    [tex]1 ^ 5 \equiv 1 \text{ mod} 10[/tex]
    2. Assume the statement is true for n = k, which means:
    [tex]k ^ 5 \equiv k \text{ mod} 10[/tex], or, equivalently, [tex](k ^ 5 - k) \vdots 10[/tex]
    3. Prove the statement is true for n = k + 1.
    (k + 1)5 - (k + 1) = (k5 + 5k4 + 10k3 + 10k2 + 5k + 1) - (k + 1)
    = (k5 - k) + (10k3 + 10k2) + (5k4 + 5k)

    From the Induction Hypothesis, k5 - k is divisible by 10, and obviously 10k3 + 10k2 is divisible by 10.

    What's left is to prove 5k4 + 5k is divisible by 10.

    ... (Write that part of your proof here)

    Hence (k + 1)5 - (k + 1) is divisible by 10, hence [tex](k + 1) ^ 5 \equiv k + 1 \text{ mod} 10[/tex] (Q.E.D)


    ---------------------------------

    Another idea is to use the product of consecutive integers to prove it. The aim is to prove x5 - x is divisible by 10, for all x, right?

    x5 - x = x(x4 - 1) = x(x2 - 1)(x2 + 1)
    = (x - 1)x(x + 1)(x2 + 1)

    Of course (x - 1)x(x + 1)(x2 + 1) is divisible by 2. (Hint: There's a product of 2 consecutive integers in that expression). So what's left is to prove that expression is also divisible by 5.

    (x - 1)x(x + 1) is already a product of 3 consecutive integer. Now, let's see if you can think of a way so that the expression (x - 1)x(x + 1)(x2 + 1) can be split into the product of 5 consecutive integers. If you can then, everything is done. :)
     
  4. Sep 29, 2009 #3
    so we just learned about Fermat's little theorem and it makes divisibility by 5 a trivial problem so I just used that. thank you for the help. I will do my best to make the presentation better
     
    Last edited: Sep 29, 2009
  5. Sep 29, 2009 #4

    Landau

    User Avatar
    Science Advisor

    Often a==b(mod n) by definition means that n divides a-b. What is your definition?
     
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