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Suppose $x + 2y = k$ where $k$ is prime and $x > y > 0$.

Prove that $x|\binom{x}{y}$. That is, prove that $\binom{x}{y}$ is divisible by $x$.

The only leads I have are that, since $x$ and $y$ must be relatively prime -- or else $k$ would not be prime -- in the expansion of the binomial coefficient:

$$\frac{(x)(x-1)\dots(x-y+1)}{(y)(y-1)\dots 1}$$, there is necessarily some value in $\{x-1 \dots x-y+1\}$ that is divisible by $y$. This is true, because given any consecutive $y$ numbers, at least one must be divisible $y$, and we know that $x$ is not that number.

Edit: Apparently this is answered here When is $\binom{n}{k}$ divisible by $n$?

I'm not seeing how the result is obvious, however, for the particular case that $(n,k) = 1$.

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Muno
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    We can write

    $$\binom{n}{k} = \frac{n}{n-k}\binom{n-1}{k}.$$

    Since $n$ and $n-k$ are coprime, and $\binom{n}{k}$ is an integer, it follows that $(n-k) \mid \binom{n-1}{k}$, and hence $$\frac{1}{n}\binom{n}{k} = \frac{1}{n-k}\binom{n-1}{k}$$

    is an integer.

     – Daniel Fischer Dec 24 '16 at 13:51

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We can use this nice identity $$ \binom{x}{y}\binom{x-y}{h} = \binom{x}{h}\binom{x-h}{y} $$ to obtain $$ \binom{x}{y}y = \binom{x}{y}\binom{y}{1} = \binom{x}{1}\binom{x-1}{x-y} = x\binom{x-1}{x-y}. $$ Now we see that $x$ divides right side, so it must divide also left side. But since $(x,y)=1$, $x$ must divide $\binom{x}{y}$.

Sil
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