Exercise 2.7.9 (Wolstenholme's congruence for $p=5$.)

Question

For $p = 5$, compute the values of the numbers $\sigma_1,\sigma_2,\sigma_3, \sigma_4$ in (2.7).

Richard Ganaye · Accepted Answer

\begin{proof} 
By (2.7) for $p = 5$
$$(x-1)(x-2)(x-3)(x- 4) = x^4 - \sigma_1 x^3 +\sigma_2 x^2 -\sigma_3 x + \sigma_4.$$
Expanding the left member, we obtain
$$x^4 - 10 x^3 + 35 x^2 - 50 x + 24 =  x^4 - \sigma_1 x^3 +\sigma_2 x^2 -\sigma_3 x + \sigma_4,$$
so
$$\sigma_1 = 10, \quad \sigma_2 = 35, \quad \sigma_3 = 50,\quad  \sigma_4 = 24.$$
Note that $\sigma_3 \equiv 0 \pmod{5^2}$: this is the Wolstenholme'congruence for $p = 5$.
\end{proof}

Exercise 2.7.9 (Wolstenholme's congruence for $p=5$.)

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Exercise 2.7.9 (Wolstenholme's congruence for $p=5$.)

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