In the book of [Z.I._Borevitch,_I.R._Chafarevitch]_Théorie_des_nombres p.21 the authors talking about the construction of 7-adic integers by considering equations of type $$(C_n): x^2\equiv 2\mod 7^n$$ for $n=1$ the solution satisfy $x_0\equiv \pm 3\mod 7$. For $n=2$, the solutions $(C_2)$ are in the solutions of $(C_1)$ so he considers those of the form $x_1=3+7t_1$, and replacing in $(C_2)$ implies \begin{align*} (3+7t_1)^2&\equiv 2\mod 7^2\\ 9+6.7t_1+7^2t_1^2&\equiv 2\mod 7^2\\ 1+6t_1&\equiv 0\mod 7\\ \end{align*} For me it is not understood why $1+6t_1\equiv 0\mod 7$ ? this is my problem.
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$$ 7 + 6 \cdot 7 t_1 + 7^2 t_1^2 \equiv 0 \pmod{7^2}, $$ $$ 7 + 6 \cdot 7 t_1 \equiv 0 \pmod{7^2}, $$ $$ 7 (1 + 6 \cdot t_1) \equiv 0 \pmod{7^2}, $$ – Will Jagy Mar 10 '23 at 22:56
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1$\begin{align} &\bmod 7\cdot\color{#c00}7!:\ {-}\color{#c00}7\equiv 6t\cdot \color{#c00}7 + \overbrace{7^2 t^2}^{\equiv\ 0}\[.2em] \Longrightarrow\ &\bmod 7!:\ \ \ ,\ \ {-}1\equiv 6t,\ \text{ by cancel $\color{#c00}7$ as in the linked dupe}\end{align}$ – Bill Dubuque Mar 10 '23 at 23:05
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Any thoughts about the comment and answers, Muhammad? – Gerry Myerson Mar 12 '23 at 12:01
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@GerryMyerson very clear thanks. – Muhammad rahmani Mar 13 '23 at 09:36
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If one of the answers was satisfactory, you can "accept" it by clicking in the check mark next to it. – Gerry Myerson Mar 13 '23 at 09:52
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As an intermediate step between your second and third displayed congruences, we have $(9-2)+6\cdot 7\cdot t_1 + 7^2\cdot t_1=0 \mod 7^2$. Since $7a=0 \mod 7^2$ implies $a=0\mod 7$ (!), we get $1 + 6t_1+7t_1^2=0\mod 7$. The $7t_1$ term is $0$ mod $7$, so this is $1+6t_1=0\mod 7$.
paul garrett
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Let me put in some missing steps. From $9+6\cdot7t_1+7^2t_1^2\equiv2\bmod{7^2}$ we subtract $2$ to get
$7+6\cdot7t_1+7^2t_1^2\equiv0\bmod{7^2}$ then we divide through by $7$ to get
$1+6t_1+7t_1^2\equiv0\bmod{7}$ and that gets you
$1+6t_1\equiv0\bmod{7}$.
Gerry Myerson
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