Exercise 2.3.1 (First examples, part 1)

Question

Find the smallest positive integer (except $x= 1$) that satisfies the following congruences simultaneously: $x \equiv 1 \pmod 3, x\equiv 1 \pmod 5, x \equiv 1 \pmod 7$.

Richard Ganaye · Accepted Answer

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

\begin{proof} Since $3,5,7$ are prime by pairs, by theorem 2.3 (part 3),
\begin{align*}
x \equiv 1 \pmod 3, x\equiv 1 \pmod 5, x \equiv 1 \pmod 7 &\iff 3 \mid x-1, 5 \mid x-1, 7 \mid x-1\
&\iff 3\cdot 5 \cdot 7 \mid x-1\
&\iff \exists k \in \Z,\ x = 1 + k 105.
\end{align*}
Therefore the smallest solution $x_1 >1$ is $x_1 = 106$.
\end{proof}

Exercise 2.3.1 (First examples, part 1)

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Exercise 2.3.1 (First examples, part 1)

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