Binomial coeff having bounded residue Let $\{x\}$  denote closest multiple of $p$ to $x$. Prove that for prime $p =4k+1$,
$$ \left | \frac{1}2 \dbinom{2k}k-\left \{\frac{1}2 \dbinom{2k}k \right \} \right|< \sqrt p.$$
 A: Let $\chi:\Bbb F_p^*\to\Bbb C^*$ be a quartic character. Then $\chi^2$
is the Legendre symbol modulo $p$. By definition the Jacobi sum
$$J(\chi,\chi^2)=\sum_{a=0}^{p-1}\chi(a)\chi^2(1-a)$$
where we follow the usual convention that $\chi(a)=0$ when $p\mid a$.
It is well-known (see for instance Ireland and Rosen's A Classical Introduction to Modern Number Theory) that $J(\chi,\chi^2)=\pi$
where $\pi$ is a Gaussian integer of norm $p$. Moreover
$$\chi(a)\equiv a^k\pmod\pi$$
in the Gaussian integers. Now consider congruences modulo $\overline\pi$.
Then
$$\chi(a)\equiv a^{3k}\pmod{\overline\pi}$$
and
$$\chi^2(a)\equiv a^{2k}\pmod{\overline\pi}.$$
Then
\begin{align}
\pi&=J(\chi,\chi^2)\equiv\sum_{a=0}^{p-1}a^{3k}(1-a)^{2k}\\
&\equiv\sum_{j=0}^{2k}\sum_{a=0}^{p-1}(-1)^j\binom{2k}ja^{3k+j}\pmod{\overline\pi}
\end{align}
A sum $\sum_{a=0}^{p-1}a^r$ is divisible by $p$ unless $r$ is
a multiple of $p-1=4k$. In the above sum all the terms in the
$j$ sum are divisible by $p$ and so also $\pi$
save for the $j=k$ term. Thus
$$\pi\equiv\sum_{a=0}^{p-1}(-1)^k\binom{2k}k a^{4k}\equiv-(-1)^k\binom{2k}
k\pmod{\overline\pi}.$$
Write $\pi=b+ci$. Then $b^2+c^2=p$ and $\pi\equiv 2b\pmod{\overline\pi}$.
Therefore
$$\binom{2k}k\equiv\pm2b\pmod{\overline\pi}$$
and as bot sides are integers then
$$\binom{2k}k\equiv\pm2b\pmod p.$$
As $\binom{2k}k$ is even, then
$$\frac12\binom{2k}k\equiv\pm b\pmod p.$$
The difference of $\frac12\binom{2k}{k}$ and the nearest multiple of $p$
is $\pm b$ and $b^2=p-c^2<p$.
