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Binomial Theorem and Modular Arithmetic Proof Check

  1. Oct 21, 2009 #1
    1. The problem statement, all variables and given/known data

    [tex] \mbox{Prove or give a counterexample: If p is a prime integer, then for all integers x and y, } (x+p)^p \equiv_p x^p+y^p[/tex].

    2. Relevant equations

    [tex]\equiv_p \mbox{just means (mod p).

    Can you please check and see if this proof is well-formed?}[/tex]

    3. The attempt at a solution

    [tex]\mbox{Pf: Assume p is prime. Then} (x+y)^p=
    \left(\begin{array}{l c}
    p\\
    0\\
    \end{array}\right)

    x^p+

    \left(\begin{array}{l c}
    p\\
    1\\
    \end{array}\right)

    x^{p-1}y+

    \left(\begin{array}{l c}
    p\\
    2\\
    \end{array}\right)

    x^{p-2}y^2+ ... +
    \left(\begin{array}{c c}
    p\\
    p-1\\
    \end{array}\right)

    xy^{p-1}+

    \left(\begin{array}{l c}
    p\\
    p\\
    \end{array}\right)
    y^p = x^p + \sum^{p-1}_{k=1}x^ky^{p-k}+y^p. [/tex]

    [tex]\mbox{Notice that} \sum^{p-1}_{k=1}x^ky^{p-k} = \frac{p!}{k!(p-k)!}\mbox{, which, by a previously proven theorem, we know that } p\left| \frac{p!}{k!(p-k)!} \right..[/tex]

    [tex]\mbox{ So, by the definition of (mod p) we know that} \sum^{p-1}_{k=1}x^ky^{p-k}+y^p \equiv_p 0 \mbox{ and therefore }(x+p)^p \equiv_p x^p+y^p ~~~~\hspace{1.0cm}_\blacksquare[/tex]
     
  2. jcsd
  3. Oct 21, 2009 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    It's not well formed. [tex]\sum^{p-1}_{k=1}x^ky^{p-k} = \frac{p!}{k!(p-k)!}[/tex] is completely false. You are summing over k. There can't be a k in the answer. The point is that each individual binomial coefficient C(p,k) is divisible by p for 1<=k<=p-1.
     
  4. Oct 21, 2009 #3
    I discovered my error. Give me a minute.
     
  5. Oct 21, 2009 #4
    [tex]
    \mbox{Pf: Assume p is prime. Then} (x+y)^p=
    \left(\begin{array}{l c}
    p\\
    0\\
    \end{array}\right)

    x^p+

    \left(\begin{array}{l c}
    p\\
    1\\
    \end{array}\right)

    x^{p-1}y+

    \left(\begin{array}{l c}
    p\\
    2\\
    \end{array}\right)

    x^{p-2}y^2+ ... +
    \left(\begin{array}{c c}
    p\\
    p-1\\
    \end{array}\right)

    xy^{p-1}+

    \left(\begin{array}{l c}
    p\\
    p\\
    \end{array}\right)
    y^p [/tex]
    [tex] = x^p + \sum^{p-1}_{k=1}\left(\begin{array}{l c}
    p\\
    k\\
    \end{array}\right)x^ky^{p-k}+y^p.
    [/tex]

    [tex]
    \mbox{Notice that} \sum^{p-1}_{k=1}\left(\begin{array}{l c}
    p\\
    k\\
    \end{array}\right)x^ky^{p-k} = \sum^{p-1}_{k=1}\left[\left(\frac{p!}{k!(p-k)!}\right) x^ky^{p-k}\right]\mbox{, which, by a previously proven theorem, we know that } p\left| \frac{p!}{k!(p-k)!} \right..
    [/tex]

    [tex]
    \mbox{ So, by the definition of (mod p) we know that} \sum^{p-1}_{k=1}\left(\begin{array}{l c}
    p\\
    k\\
    \end{array}\right)x^ky^{p-k}+y^p \equiv_p 0 \mbox{ and therefore }(x+y)^p \equiv_p x^p+y^p ~~~~\hspace{1.0cm}_\blacksquare
    [/tex]
     
    Last edited: Oct 21, 2009
  6. Oct 21, 2009 #5
    Alright. How does that look?
     
  7. Oct 21, 2009 #6

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    The third line is still completely garbled. The second line would say you can ignore the k=1 to k=p-1 terms because they are all divisible by p. Say the the previous theorem you are citing is for 1<=k<=p-1. On the last line (x+p)^p? Come on.
     
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