How can I find the remainder when $\displaystyle\binom{2018}{1009}$ is divided by $ 2017^2?$
My Approach:
I have used Lucas' theorem Extension here,that is
$$\binom{n}{m} \equiv \frac{P}{Q}(\mod p^f)$$
$$\text{where},P=\prod_{i=0}^{s-f+1} \binom{n_i+n_{i+1}p+ \dots+n_{i+f-1}p^{f-1}}{m_i + m_{i+1}p+ \dots+ m_{i+f-1}p^{f-1}}$$
$$\& \space Q=\prod_{i=1}^{s-f+1} \binom{n_i+n_{i+1}p+ \dots+n_{i+f-2}p^{f-2}}{m_i + m_{i+1}p+ \dots+ m_{i+f-2}p^{f-2}}$$
so applying Lucas' theorem Extension in this problem:
$$\text{We can write : } \binom{2018}{1009}= \binom{1 \times 2017 \quad \quad 1 \times 2017^0}{0 \times 2017 \quad \quad 1009 \times 2017^0}$$
$$\implies s=1,f=2$$
$$\therefore P= \prod_{i=0}^{0} \binom{n_i+n_{i+1}p}{m_i + m_{i+1}p} \space \& \space Q= \prod_{i=1}^{0} \binom{n_i}{m_i }$$
$$\implies P= \binom{1 \times 2017 \quad \quad 1 \times 2017^0}{0 \times 2017 \quad \quad 1009 \times 2017^0} \space \& \space Q= 1$$
so again we are getting:
$$\binom{2018}{1009}=\frac{\binom{2018}{1009}}{1} \mod (2017^2)$$
Thus, ${\color{red}{\text{NO IMPROVEMENT}}}$
so how can i evaluate its remainder??
and if possible then how can we modify this process to get the answer? please help...
Note that
$$\binom{2018}{1009}=\frac{2018!}{1009!^2}=2018\cdot 2017\frac{2016!}{1009!^2}\equiv x \pmod{2017^2}\\\iff 2018\cdot \frac{2016!}{1009!^2}\equiv \frac{x}{2017} \pmod{2017}$$
and since
$$(p-1)!\equiv -1 \pmod{p}$$
$$\left[\left(\frac{p-1}{2}\right)!\right]^{2}\equiv (-1)^{\frac{p+1}{2}}\pmod{p}$$
we have that $$ \frac{x}{2017}\equiv2018\frac{2016!}{1009!^2} \equiv \frac{2016!}{1009^2\cdot1008!^2}\equiv \frac{1}{1009^2} \iff \frac{x\cdot 1009^2}{2017}\equiv 1 \pmod{2017}$$
note that
$$1009^2=\left( \frac{2017+1}{2}\right)^2=\frac14(2017^2+2\cdot2017+1)\equiv\frac14 \pmod{2017}$$
thus
$$\frac{x\cdot 1009^2}{2017}\equiv 1 \iff \frac{x}{4\cdot 2017}\equiv 1 \pmod{2017}\iff x \equiv 8068 \pmod{2017^2}$$