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If $a\equiv d\mod r$, $a\equiv e\mod s$, $r$ and $s$ are coprime and $m=rs$, then we have to show that $a\equiv \gcd(d,e)\mod m$

My try: By applying Chinese Remainder Theorem finally I g0t $a\equiv(1+rt_1)d+(1+st_2)e\,mod\,m$.Can I conclude $(1+rt_1)d+(1+st_2)e$ as gcd $(d,e)?$

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    When you type want to write a modulus, you can just do `\mod n`, you don't need to bother with `\,mod\,n`. – 2017-02-24

1 Answers 1

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The assertion is wrong. Consider

$$a \equiv 2 \pmod 3, a \equiv 4 \pmod 5$$

The solution to this system is $a \equiv 14 \pmod {15}$, not $a \equiv 2 \pmod {15}$.

Another way to see this is that you could reformulate $a \equiv 2 \pmod 3$ as $a \equiv 5 \pmod 3$, and suddenly your assertion changes from $a \equiv 2 \pmod {15}$ to $a \equiv 1 \pmod {15}$.