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Let $p_1$ and $p_2$ be distinct primes. Is it always true that the solution of the simultaneous congruence $$ax \equiv 1 \mod{p_1}$$ $$ax \equiv 1 \mod{p_2}$$ is a solution to $ax \equiv 1 \mod{p_1p_2}?$

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I think yes. If $p_1 \mid (aX-1)$ and $p_2 \mid (aX-1)$ then $p_1 p_2=lcm(p_1,p_2) \mid (aX-1)$.

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  • $\begingroup$ I think that the greatest common divisor of two primes is 1, not p1*p2. Did you mean LCM? $\endgroup$ – Adam Oct 6 '17 at 11:37
  • $\begingroup$ Oh yes, you are correct! I am going to edit my answer. $\endgroup$ – Huy Dang Oct 6 '17 at 12:12

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