- #1
Hotsuma
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Homework Statement
[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].
Homework Equations
[tex]\equiv_p \mbox{just means (mod p).
Can you please check and see if this proof is well-formed?}[/tex]
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]