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Let $a$ and $b$ be two integers greater than 1. Show that there exists an odd integer $x$ such that $x \equiv a \mod {b}$ and $x \equiv b \mod {a}$

Let $a$ and $b$ be two integers greater than 1. Show that there exists an odd integer $x$ such that $x \equiv a \mod {b}$ and $x \equiv b \mod {a}$

The outline of the proof, so far:
If $a=2$, then let $x=2$.
If $a > 2$, assume by contradiction that $x \equiv a \mod {b}$ and $x \equiv b \mod {a}$ for all odd $x$, then there exists a prime $p$ such that $\frac ab = \frac x {p-1}$. Since $a$ and $b$ are greater than 1, we have $x$ is odd. Therefore, $\frac x {p-1}$ is even. But this is impossible, since an even number divided by an odd number gives an odd number.
In the proof above, can this be simplified? I’ve tried this but it doesn’t seem to work. Thanks.

A:

As the comments above mention, you have a few issues.
First, you know that $a$ and $b$ are greater than $1$. (You don’t need that fact.) However, it is
$$x=\frac{ab}{\gcd(a,b)}
c6a93da74d

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