Exercise 2.7.14* (Mysterious congruence)

Proof. First estimation of $a$.

Write ${[a]}_p$ the class of $a$ in $\mathbb{Z}∕\mathit{p\mathbb{Z}}$. By definition of $a$,

By Wilson’s theorem, $(p-1)!\equiv -1(\mathit{mod}p)$, thus

Moreover $2^p={(1+1)}^p=2+{\sum <!--\; FUNCTION\; APPLICATION\; -->}_{k=1}^{p-1}\left(\genfrac{}{}{0.0pt}{}{p}{k}\right)$. If we write $b=(2^p-2)∕p$, by Fermat’s theorem, $b$ is an integer, and

Therefore, using $k!\not\equiv 0$ for all $k\in [[1,p-1]]$,

We obtain ${[b]}_p=-{[a]}_p$, so $a\equiv \frac{2-2^p}{p}(\mathit{mod}p)$. The solution is done. To conclude,

(By problem (10), we have also

Second estimation of $a$.

Let $g(x)\in \mathbb{Z}[x]$ be the polynomial given by

whose roots are

We write, as in the proof of Wolstenholme’s congruence,

Here

Moreover, by definition of $a$,

Then

The idea is to compute two evaluations of $g(p)$ modulo $p^2$, to obtain ${\tau }_{p-2}$ modulo $p$.

• First

  $$\begin{array}{llll}\hfill g(p)& =p^{p-1}-{\tau }_1{p}^{p-1}+\cdots +{\tau }_{p-3}{p}^2-{\tau }_{p-2}p+{\tau }_{p-1},& \hfill & \\ \hfill & & \hfill \end{array}$$

  therefore

  $$\begin{array}{lll}\hfill g(p)\equiv -{\tau }_{p-2}p+{\tau }_{p-1}(\mathit{mod}{p}^2).& & \hfill \text{(1)}\end{array}$$

• Next

  $$\begin{array}{llll}\hfill g(p)& =(p-1)(p+2)(p-3)\cdots (p-(p-2))(p+(p-1))& \hfill & \\ \hfill & =[(p-1)(p-3)\cdots 2]\cdot [(p+2)(p+4)\cdots (p+(p+1))& \hfill & \\ \hfill & =2^{(p-1)∕2}\left[\frac{p-1}{2}\frac{p-3}{2}\cdots 1\right]\frac{(p+1)(p+2)(p+3)(p+4)\cdots (p+p)}{(p+1)(p+3)\cdots (p+p)}& \hfill & \\ \hfill & =2^{(p-1)∕2}\left(\frac{p-1}{2}\right)!\frac{(2p)!}{p!}\frac{1}{2^{(p+1)∕2}\left(\frac{p+1}{2}\right)\left(\frac{p+3}{2}\right)\cdots p}& \hfill & \\ \hfill & =\frac{1}{2}\left(\frac{p-1}{2}\right)!\frac{(2p)!}{p!}\frac{\left(\frac{p-1}{2}\right)!}{p!}& \hfill & \\ \hfill & =\frac{1}{2}\left(\genfrac{}{}{0.0pt}{}{2p}{p}\right){\left[\left(\frac{p-1}{2}!\right)\right]}^2& \hfill & \\ \hfill & =\left(\genfrac{}{}{0.0pt}{}{2p-1}{p-1}\right){\left[\left(\frac{p-1}{2}\right)!\right]}^2& \hfill & \end{array}$$

  By Problem 12, we know that $\left(\genfrac{}{}{0.0pt}{}{2p-1}{p-1}\right)\equiv 1(\mathit{mod}{p}^2)$ (even if $p=3$), thus

  $$\begin{array}{lll}\hfill g(p)\equiv {\left[\left(\frac{p-1}{2}\right)!\right]}^2(\mathit{mod}{p}^2),& & \hfill \text{(2)}\end{array}$$

  where we saw in the solution of exercise 2.1.18 that

  $$\begin{array}{lll}\hfill {\left[\left(\frac{p-1}{2}\right)!\right]}^2\equiv {(-1)}^{(p+1)∕2}(\mathit{mod}p).& & \hfill \text{(3)}\end{array}$$

  Then (1) and (2) give

  $$-{\tau }_{p-2}p+{\tau }_{p-1}\equiv {\left[\left(\frac{p-1}{2}\right)!\right]}^2(\mathit{mod}{p}^2),$$

  that is

  $$-{(-1)}^{(p-1)∕2}ap+{(-1)}^{(p-1)∕2}(p-1)!\equiv {\left[\left(\frac{p-1}{2}\right)!\right]}^2(\mathit{mod}{p}^2).$$

  Therefore

  $$a\equiv \frac{(p-1)!-{(-1)}^{(p-1)∕2}{\left[\left(\frac{p-1}{2}\right)!\right]}^2}{p}(\mathit{mod}p),$$

  where the numerator is divisible by $p$, by(3) and Wilson’s theorem.

  This is not the expected result. If we compare the two results, we obtain as promised the smart congruence, true for every odd prime,

  $$\frac{(p-1)!-{(-1)}^{(p-1)∕2}{\left[\left(\frac{p-1}{2}\right)!\right]}^2}{p}\equiv \frac{2-2^p}{p}(\mathit{mod}p),$$

  (or equivalently $(p-1)!-{(-1)}^{(p-1)∕2}{\left[\left(\frac{p-1}{2}\right)!\right]}^2\equiv 2-2^p(\mathit{mod}{p}^2)$.)