$p_1, p_2$ are distinct prime numbers
I have just observed this pattern when solving this problem. Is there a simple way to prove/disprove it ?
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You need them to be distinct primes... – Thomas Andrews Sep 17 '14 at 15:21
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yes otherwise x=a mod p^2 would not make sense, I'll edit the question :) – AgentS Sep 17 '14 at 15:22
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2I wouldn't say it does "not make sense," I would say it's false. Consider $x = 13$, $a = 1$, and $p_1 = p_2 = 3$, then $13 \equiv 1 \mod 3$ but $13 \equiv 4 \mod 3^2$. – Sep 17 '14 at 15:49
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More generally, if $x\equiv a\pmod m$ and $x\equiv a\pmod n$ then $x\equiv a\pmod{\mathrm{lcm}(m,n)}$.
This is true because $x-a$ is a multiple of $m$ and a multiple of $n$, so must be a multiple of the least common multiple of $m,n$.
If $\gcd(m,n)=1$ then $\mathrm{lcm}(m.n)=mn$.
And if $p_1,p_2$ are distinct primes, then $\gcd(p_1,p_2)=1$.
Thomas Andrews
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wow! that's perfect explanation, goes straight into my head thank you so much (: – AgentS Sep 17 '14 at 15:25
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1If you are answering a question for a class, some of these assertions require proof. For example, the fact that any common multiple is a multiple of the least common multiple actually requires proof. (It's not hard.) Similarly, the assertion about the case $\gcd(m,n)=1$ requires proof. Even the fact that two distinct primes are relatively prime technically requires proof. – Thomas Andrews Sep 17 '14 at 15:33