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I'm not sure how to do this:

show that $(x + 12y)^{13}\equiv x^{13} - y^{13} {\mod 13}$ for all integers

It's part of a question on the binomial theorem, but that's all I've got.

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2 Answers 2

Hint 1: $(x + 12y)^{13}\equiv (x - y)^{13} {\mod 13}$

Hint 2: in the expansion of $(a + b)^{p}$ for $p$ prime, each term has a co-efficient which is a multiple of $p$ except the first and last.

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2  
Why is hint 1 true? And how do I know 2? And how do I use these facts? –  twok0boo Jan 23 '12 at 8:48
    
I don't know what you know. Hint 1 is true because $13y \equiv 0 {\mod 13}$ and because $(a+b c)^d \equiv a^d {\mod b}$. You use these facts by letting $a=x$, $b=-y$ and $p=13$. –  Henry Jan 23 '12 at 8:54

Hint :

If we apply freshman dream theorem we can write :

$(x+12y)^{13} \equiv x^{13}+(12y)^{13} \pmod {13}$

Now , use a fact :

$12 \equiv -1 \pmod {13} \Rightarrow 12^{13} \equiv -1 \pmod{13}$

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