Hi !

I've got a little problem, seems simple... Here it is :

Text :

We're going to study all couples (n,k) such as :

$\displaystyle {n \choose k}={n+1 \choose k-1}$

Question :

Let $\displaystyle p=n-k+1$ and $\displaystyle t=gcd(p,k)$. Also $\displaystyle s=\frac{k}{t}$

Show that $\displaystyle p=t^2$

So far... :

By replacing the combinations with their formulas with !, i've found that :

$\displaystyle p(p+1)=(p+k)k$

$\displaystyle p^2+p=pk+k^2$

So it's easy to show that $\displaystyle t^2|p$

And now... ?

I can't find how to show that $\displaystyle p|t^2$ or $\displaystyle p \leq t^2$ :'(

I thought Bézout's identity could help, but no...

Thanks in advance !