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I am following a proof of an RSA algorithm and the proof states the following:

$p_1$ and $p_2$ are distinct primes, $a \equiv b \pmod{p_1}$ and $a \equiv b \pmod{p_2} \Rightarrow a \equiv b \pmod{p_1\times p_2}$. Let's call that FACT X.

can someone give me a proof for this. i have no formal mathematical training so a detailed proof will be much appreciated.

Ishan Antony
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  • You need to clarify the nature of the $p_i$. Are they, say, distinct primes? (the claim is false if $p_1=p_2$) – lulu May 30 '16 at 12:39
  • yes they are distinct primes – Ishan Antony May 30 '16 at 13:24
  • That should appear in the question, not just a comment. I'll post something below. – lulu May 30 '16 at 13:28
  • thanks mentioned it in the question – Ishan Antony May 30 '16 at 13:30
  • This follows from the chinese remainder theorem I think. – M. Van May 30 '16 at 13:37
  • @M.Van, yes it was mentioned in the proof but it is difficult for me to understand, would you mind explaining it? – Ishan Antony May 30 '16 at 13:42
  • You say you have no formal mathematical training. I can explain it easily if you know about bijections and about $\mathbb{Z}_n$, the group of residue classes mod n. Do you know these concepts? – M. Van May 30 '16 at 13:47
  • i understand bijection, one to one and onto when we are talking about bijective functions. From what you mention about Zn, i have a feeling that Zn is the set of possible values created by mod n, the integers 1 to n - 1 am i correct? – Ishan Antony May 30 '16 at 13:52

2 Answers2

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We note that $$p_1|(a-b)\implies a-b=p_1m$$ for some integer $m$.

But then we have $p_2|p_1m\implies p_2|m$ (if a prime divides a product then it divides at least one of the factors and we know that $p_2$ does not divide $p_1$). Thus $m=p_2n$ for some integer $n$, whence $a-b=p_1p_2n$ so $p_1p_2|a-b$, as desired.

lulu
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Here is the key fact:

If $u$ and $v$ divide $w$, then $lcm(u,v)$ divides $w$.

Apply this to $u=p_1$, $v=p_2$ and $w=a-b$, noting that $lcm(p_1,p_2)=p_1 p_2$ because $p_1\ne p_2$.

lhf
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