How to prove $p^2 \mid \binom {2p} {p }-2$ for prime $p$?

Question

How to prove $p^2 \mid \binom {2p} {p } -2$ for prime $p$?

I have a hint: for $1 \le i \le p-1$, $p \mid \binom p i$.

I cannot even start the proof. Please help.

Answer 1

Note that: $$\binom{2p}p=\sum_{i\mathop=0}^p\binom pi\binom p{p-i}=\sum_{i\mathop=0}^p\binom pi^2$$

In fact, this formula can be generalized:

$$\binom{a+b}r=\sum_{i\mathop=0}^r\binom ai\binom b{r-i}$$

The proof can be derived by considering the coefficient of $x^r$ in the expansion of $(1+x)^{a+b}$ and $(1+x)^a(1+x)^b$ respectively, which would be the same.

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Now, from your hint, since $p$ is actually a prime, we have: $$p^2|\tbinom pi^2$$ where $1\le i\le p-1$.

Therefore: $$\sum_{i\mathop=0}^p\binom pi^2=\dbinom p0^2+\sum_{i\mathop=1}^{p-1}\binom pi^2+\binom pp^2\equiv\dbinom p0^2+0+\dbinom pp^2=2\mbox{ (mod p}^2\mbox{)}$$

Answer 2

Hint: $$ \binom{2p}{p}=\sum_{i=0}^p\binom{p}{i}\binom{p}{p-i}. $$ This is because choosing $p$ elements from $2p$ elements is the same as choosing $i$ from the first $p$ and $p-i$ from the last $p$ elements. Now, you can use the hint in your statement to finish the proof.