Exercise 1.28

Question

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For all $n$ show that $30 \mid  n^5 - n$ and that $42\mid n^7 - n$.

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

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\begin{proof}
If we want to avoid Fermat's Little Theorem (Prop. 3.3.2. Corollary 2 P. 33), note that
\begin{align*}
(n-2)(n-1)n(n+1)(n+2) &= n(n^2-1)(n^2-4)\
&=n^5-5n^2+4n\
&=n^5 - n +5 (-n^2+n)
\end{align*}
As the product of 5 consecutive numbers is divisible by 5,
$$5 \mid n^5 - n.$$
Moreover,
\begin{align*}
&2 \mid (n-1)n \mid (n^4-1)n = n^5 -n\
&3 \mid (n-1)n(n+1) = n(n^2 - 1) \mid n(n^4-1) = n^5 - n
\end{align*}
As $2,3,5$ are distinct primes, $2	imes 3 	imes 5 = 30 \mid n^5-n$.

Similarly,
\begin{align*}
(n-3)(n-2)(n-1)n(n+1)(n+2)(n+3) &= n(n^2-1)(n^2-4)(n^2-9)\
&=n(n^4-5n^2+4)(n^2-9)\
&=n^7-14n^5+49n^3-36n\
&=n^7-n + 7(-2n^5+7n^3-5n)
\end{align*}
As the product of 7 consecutive numbers is divisible by 7,
$$7 \mid n^7 -n.$$
Moreover
\begin{align*}
&2 \mid (n-1)n  \mid (n^6 - 1)n= n^7 -n\
&3 \mid (n-1)n(n+1) = n(n^2 - 1) \mid n(n^6-1) = n^7 - n
\end{align*}
As $2,3,7$ are distinct primes $2	imes 3 	imes 7 = 42 \mid n^7-n$.
\end{proof}

Exercise 1.28

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Exercise 1.28

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