Exercise 5.6.12 (Value of $
u_7(Z_n)$, where $X_n^3 + Y_n^3 = 7Z_n^3$)

Question

Let the triple $(X_n, Y_n, Z_n)$ of integers be defined as in the proof of Theorem 5.16. Show that the power of $7$ dividing $Z_n$ tends to infinity with $n$.

Richard Ganaye · Accepted Answer

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

\begin{proof} 
For some unknown reason, this result seems false, because $Z_n 
ot \equiv 0 \pmod {7^2}$ for any value of $n$.
\begin{center}
\begin{tabular}{|c|ccc|}
\hline
$n$ & $X_n$ & $Y_n$ & $Z_n$\
\hline
0 & 2 & 1 & 1\
1 & 20 & 32  & 7\
2 &  34& 25  & 35 \
 3&41 &46& 28  \
 4&  20 & 32& 42 \
 5&  34&25& 14 \
 6&  41& 46& 21  \
 7&  20& 32& 7 \
\hline
\end{tabular}
\end{center}
Thus 
$$X_7 \equiv X_1,\qquad Y_7 \equiv Y_1,  \qquad Z_7 \equiv Z_1 \pmod {49}$$
By induction, for all $n\geq 1$,
$$X_{n+6} \equiv X_n, \qquad Y_{n+6} \equiv Y_n,  \qquad Z_{n+6} \equiv Z_n \pmod {49},$$
therefore, for any $n\geq 1$ and for any $k\geq 0$
$$X_{n+6k} \equiv X_n, \qquad Y_{n+6k} \equiv Y_n,  \qquad Z_{n+6k} \equiv Z_n \pmod {49}.$$
So every $Z_n$ is congruent modulo $7^2$ to some $Z_r$, where $1 \leq r \leq 6$. By the preceding array, $Z_r 
ot \equiv 0 \pmod {7^2}$ for $1\leq r \leq 6$, but $Z_r \equiv 0 \pmod 7$. Therefore
$$
u_7(Z_n) = 1 \qquad (n\geq 1).$$
This contradicts the statement.
\end{proof}
With Sagemath:
\begin{verbatim}
def triple_mod(n):
    x, y, z = 2, 1, 1
    for i in range(n):
        x, y, z = (x * (x^3 + 2 * y^3)) % (7^2),
                  (-y * (2 * x^3 + y^3)) % (7^2), 
                  (z * (x^3 - y^3)) % (7^2)
    return (x,y,z)

print([triple_mod(n) for n in range(8)])
    [(2, 1, 1), (20, 32, 7), (34, 25, 35), (41, 46, 28), (20, 32, 42), 
           (34, 25, 14), (41, 46, 21), (20, 32, 7)]
\end{verbatim}

Exercise 5.6.12 (Value of $\nu_7(Z_n)$, where $X_n^3 + Y_n^3 = 7Z_n^3$)

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$n$	$X_{n}$	$Y_{n}$	$Z_{n}$

0	2	1	1
1	20	32	7
2	34	25	35
3	41	46	28
4	20	32	42
5	34	25	14
6	41	46	21
7	20	32	7

Exercise 5.6.12 (Value of $\nu_7(Z_n)$, where $X_n^3 + Y_n^3 = 7Z_n^3$)

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