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I am supposed to prove that $x = y \mod (pq) \iff x = y \mod p$ and $x = y \mod q$ with $p$ and $q$ are prime numbers. It somewhat sounds reasonable to me, but unfortunately I don't have any clue how to prove it.

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  • $\begingroup$ I'm voting to close this question as off-topic because it is about general mathematics. $\endgroup$
    – fkraiem
    Feb 21, 2017 at 4:27
  • $\begingroup$ The question is related to RSA and it's been quite useful for me since it was the last step I didn't understand from the MIT video: youtube.com/watch?v=9TNI2wHmaeI $\endgroup$ Jun 2, 2018 at 17:23

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That the left hand side implies the right hand side is easy, it follows from the following fact:

$$x\equiv y \mod{pq}\ \Leftrightarrow pq \mbox{ divides }x-y \Rightarrow p \mbox{ divides } x-y \mbox{ and } q \mbox{ divides }x-y$$

And so $x\equiv y \mod{p}$ and $x\equiv y \mod{q}$

Now, the point here is argument can be "reversed" in the case that $p$ and $q$ are relatively primes (and actually they are since we are assuming that they are different, if they're not, we can't obtain the result), it is, if $x\equiv y \mod{p}$ and $x\equiv y \mod{q}$, then p $\mbox{ divides } x-y$ and $q \mbox{ divides }x-y$, and as $p$ and $q$ are relatively prime (this is a well known theorem in number theory) we have that $pq$ divides $x-y$, thus $x\equiv y \mod{pq}$.

This is the general version of theorem I've wrote before:

Let $a,b,n\in \mathbb{Z}$ and suppose that $a$ divides $n$ and $b$ divides $n$, then $ab$ divides $n\cdot \mbox{gcd}(a,b)$. This, in the case that $a,b$ are relatively prime, gives the result i'm reffering to (because in that case $\mbox{gcd}(a,b)=1$).

Here's a small proof of this fact. Let $n=as$ and $n=bt$, write $\mbox{gcd}(a,b)=g$, we know that there exists $x,y$ integers such that $ax+by=g$ (see this post), then, multiplying this expression by $n$, we obtain $n(ax)+n(by)=ng$, replacing adequately, we have $(bt)(ax)+(as)(by)=ng$ and rearranging, we finally obtain $(ab)(xt+ys)=ng$ so $ab$ divides $ng=n\cdot \mbox{gcd}(a,b)$.

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Recall the definition for when two numbers are congruent modulo some number.

Now, prove to yourself that if two numbers are congruent modulo some product, they are also congruent modulo the factors of the product.

Next, prove to yourself that — if two numbers are congruent modulo two different moduli, then they are congruent modulo the least common multiple of the moduli.

Finally, apply these results… and you are done.

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