Exercise 2.1.17 ($61! + 1 \equiv 63! + 1 \equiv 0 \pmod{71}$)

Question

Show that $61! + 1 \equiv 63! + 1 \equiv 0 \pmod{71}$.

Richard Ganaye · Accepted Answer

\begin{proof} By Wilson's theorem, modulo $71$
\begin{align*}
-1 &\equiv 70! \
&\equiv 63 ! \cdot 64 \cdot 65 \cdot 66 \cdot 67 \cdot 68 \cdot 69 \cdot 70\
&\equiv 63! (-7) \cdot (-6) \cdot (-5) \cdot (-4) \cdot (-3) \cdot (-2) \cdot (-1)\
&\equiv -63! \, 7! \pmod{71}.
\end{align*}
Since $7! = 5040 = 70 \cdot 72 =  71^2 - 1 \equiv -1 \pmod {71}$,
$$-1 \equiv 63! \pmod{71}.$$
Moreover, $63 \cdot 62 \equiv (-8)\cdot (-9) \equiv 72 \equiv 1 \pmod {71}$, therefore
$$-1 \equiv 63! \equiv 61! \cdot 62 \cdot 63 \equiv 61! \pmod{71}.$$
This proves that
$$61! + 1 \equiv 63! + 1 \equiv 0 \pmod{71}.$$
\end{proof}

Exercise 2.1.17 ($61! + 1 \equiv 63! + 1 \equiv 0 \pmod{71}$)

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Exercise 2.1.17 ($61! + 1 \equiv 63! + 1 \equiv 0 \pmod{71}$)

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