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Let $p$ be an odd prime, and denote $$f(x)=\sum_{k=0}^{p-1}\binom{2k}{k}^2x^k.$$ Prove that for every $x\in \mathbf Z$,$$(-1)^\frac{p-1}2f(x)\equiv f\left(\frac{1}{16}-x\right)\pmod{p^2}.$$ This is a contest question, I do not know how to prove it. Thank you.

Addition: I find it's equivalent to prove that: $$\sum _{k=r}^{p-1} \frac{(-1)^r \binom{2 k}{k}^2 \binom{k}{r}}{16^{k-r}}\equiv(-1)^{\frac{p-1}{2}} \binom{2 r}{r}^2 \pmod {p^2}\tag1$$ for $r=0,1,2,\cdots p-1.$

And $(1)$ is equivalent to $$\sum _{k=r}^{p-1} \binom{2 k}{k}^2 \binom{k}{r}16^{-k} \equiv(-1)^{\frac{p-1}{2}} (-16)^{-r}\binom{2 r}{r}^2 \pmod {p^2}.$$

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What contest is it from? – Hagen von Eitzen May 19 '13 at 11:10
@Hagen von Eitzen An on-line contest initiated by individuals,in China. – Next May 19 '13 at 11:17
@Hecke Note that $\binom{2k}{k}^2$ is divisible by the squares of every prime between $k$ and $2k$, thus you need only consider the coefficients up to $\binom{2(\frac{p-1}{2})}{(\frac{p-1}{2})}$ as all coefficients past that will be divisible by $p^2$. – Ethan May 27 '13 at 6:05

Here, I think, is an idea in the right direction. Consider the formal power series expansion

$$F(x) := \frac{1}{\sqrt{1-4x}} = \sum_{k=0}^\infty {2k \choose k} x^k.$$

View this identity in $\mathcal O[[x]]$, the ring of formal power series with coefficients in the ring of integers $\mathcal O$ of $\overline{\mathbf Q_p}$. Let $i \in \mathcal O$ denote a square root of $-1$. Multiplying both numerator and denominator of $$\frac{1}{\sqrt{1-4x}}$$ by $i$, we have

$$F(x) = \frac{i}{\sqrt{4x-1}} = \frac{i}{\sqrt{1-4(\frac12-x)}} = iF(\frac12-x)$$

Reducing modulo the maximal ideal of $\mathcal O$, and using the fact that the central binomial coefficients are eventually divisible by $p$, we find that $$-\overline i g(x) =g\left(\frac12 - x\right) \text{ in }\overline{\mathbf F_p}$$

where $$g(x) = \sum_{k=0}^{2p-1}{2k \choose k} x^k.$$

This isn't what the question asks, but it's similar enough in appearance to lead me to believe that this is not a coincidence!

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