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Is $$ f(n,x,y)=\sum^{n-1}_{k=1}{n\choose k}x^{n-k}y^k,\qquad\qquad\forall~n>0~\text{and}~x,y\in\mathbb{Z}$$ always divisible by $2$?

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5  
Having $100$% $\LaTeX$ on the title isn't a good idea. –  Git Gud May 24 '13 at 6:35

4 Answers 4

(Hint) An odd number raised to any power is odd, and an even number raised to any power is even. In particular, $$ (x + y)^n \equiv (x + y) \pmod 2 $$ Using this along with the binomial formula, you should be able to prove the result.

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Thank you, Goos. –  Trancot May 24 '13 at 6:43
    
Yes, but your congruence is weak. It should be $(x+y)^n\equiv (x^n+y^n)\pmod{2}$, correct? –  Trancot May 24 '13 at 6:49
    
@BarisaBarukh that is true as well (in fact that is true for any prime). However, in the particular case of mod 2, I usually just replace all $n$th powers with the number itself, because it makes things simpler. –  Goos May 24 '13 at 6:56
    
But you are correct that using $(x + y)^n \equiv x^n + y^n$ makes the proof more direct. –  Goos May 24 '13 at 6:58

Hint: Recall binomial formula $$ (x+y)^n=\sum\limits_{k=0}^n{n\choose k} x^{n-k} y^k $$

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Yes, I know that. This is how I came to the above result. –  Trancot May 24 '13 at 6:37
    
Ok do you know anything about Frobenius automorphism? –  userNaN May 24 '13 at 6:40
    
I've used Frobenius in an elementary differential equations course. –  Trancot May 24 '13 at 6:41
    
I might be able to manage if you can explain it clearly. –  Trancot May 24 '13 at 6:42
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bin-o-mail? That sounds like an automatic mail deleter! :-) –  Asaf Karagila May 24 '13 at 6:43

In fact, Goos and Norbert have given the answer. (And you should also assume $n\in \mathbb{N}$)

$$ f(n,x,y) = (x+y)^n - x^n -y^n $$

If

  • both $x$ and $y$ are even: even - even - even = even;

  • both $x$ and $y$ are odd: even - odd - odd = even;

  • $x$ is even and $y$ is odd: odd - even - odd = even;

  • $y$ is odd and $x$ is even: just like the case above.

So, $f(n,x,y)$ is always even.

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Try setting $y=-x+a$ and expand the powers.

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