<s>[INST] Let p be prime. Prove that: Let p be prime:
$p^2\choose p$ is congruent to p (mod $p^2$) and $2p\choose p$ is congruent to 2(mod$p^2)$
I know that when p is prime p|$p\choose k$ where $p\choose k$ can be defined as $p!/k!(p-k)!$ and have tried playing around from there but cannot seem to make it work. Any help would be greatly appreciated!
 [/INST] Doing the case for $p=2$ can be done by hand, so I assume $p$ is odd.
$\binom{p^2}{p}=\frac{1\cdot 2 \cdot 3\dots p^2}{(1\cdot 2 \cdot 3 \cdot p)(1\cdot 2\cdot 3\dots \cdot p^2-p)}=\frac{(p^2-p+1)(p^2-p+2)\dots p^2}{1\cdot 2\cdot \dots p}$ If we cancel all of the factors of $p$ we get a complete residue class mod $p^2$ on the top and on the bottom we get a complete residue class, except instead of having the last $p-1$ terms we have their negatives. Luckily $p-1$ is even, so the product is the same as the complete residue class.
For $\binom{2p}{p}$ we just write $\frac{1\cdot2\cdot\dots 2p}{(1\cdot2\cdot3\dots\cdot p)(1\cdot2\cdot3\dots\cdot p)}$ cancelling $p$ with $p$ and $2p$ with $p$ we get $2\frac{1\cdot 2\cdot \dots (p-1) (p+1)\dots (2p-1)}{(1\cdot2\cdot3\dots\cdot p-1)(1\cdot2\cdot3\dots\cdot p-1)}$. Since we have to complete residue classes at both the top and bottom we get this is congruent to $2$.
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