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Prove that if $d$ and $e$ are relatively prime and $a$ is an integer, then \begin{align*}x & \equiv a \pmod{d}\\x &\equiv a \pmod{e}\end{align*} implies $x \equiv a \pmod{de}$.

The two given congruences imply $x = (a+dk_1) = (a+ek_2)$. How do we use this to prove the statement?

user19405892
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    You have $d \mid (x-a)$ and $e \mid (x-a)$. Thus $lcm (e,d) \mid (x-a)$. But the latter is $ed$ under your assumption. I think this ought to have been ask many times already. – quid Sep 08 '16 at 18:51
  • @quid How do $d \mid (x-a)$ and $e \mid (x-a)$ imply $\text{lcm}(e,d) \mid (x-a)$? – user19405892 Sep 08 '16 at 18:54
  • continue.... $dk_1 - ek_2 = 0$ so $dk_1 = e_k_2$. So $dk_1/e$ is a whole number but as $d$ and $e$ are relatively prime and have no factors in common, $k_1/e = k_3$ is a whole number. So $x = a + dek_3$. and $x \equiv a \mod de$. – fleablood Sep 08 '16 at 18:58
  • If a|c and b|c then $\frac{ab}{\gcd(a,b)}|c$ and $\lcm(a,b) = \frac{ab}{\gcd(a,b)}$. Consider the prime factors and powers of $a$ and $b$. They each divide c, so the least common multiples of the prime factors and powers must also divide c. – fleablood Sep 08 '16 at 19:06
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    $(x-a)$ is a multiple of both $d$ and $e$ thus it is a common multiple of $e$ and $d$. Every common multiple of $e$ and $d$ is a multiple of the least common multiple, almost by definition. – quid Sep 08 '16 at 19:09
  • @quid d'oh. That is sooo much more straightforward than what I said. – fleablood Sep 08 '16 at 19:15
  • Indeed quid's comment points out that this Question resolves the current one if $x-a$ is substituted for $x$. – hardmath Sep 08 '16 at 19:20
  • $d,e\mid n,\Rightarrow, {\rm lcm}(d,e)\mid n.\ $ Yours is special case $,n = x-a,$ and $,\gcd(d,e) = 1,,$ so $,{\rm lcm}(d,e) = de.\ $ – Bill Dubuque Oct 21 '16 at 03:24

2 Answers2

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$$(a+dk_1) = (a+ek_2) \Rightarrow dk_1 = ek_2.$$ Since $d$ and $e$ are coprime, then $k_1 = e q$ and $k_2 = d q$ for some $q$. This means that:

$$x = a+dk_1 = a + deq.$$

the_candyman
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continue.... $dk_1 - ek_2 = 0$ so $dk_1 = ek_2$. So $dk_1/e$ is a whole number but as $d$ and $e$ are relatively prime and have no factors in common, $k_1/e = k_3$ is a whole number. So $x = a + dek_3$. and $x \equiv a \mod de$.

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postscript: Let $e = \prod p_i^{a_i}; p_i$ prime. Let $d \prod q_i^{b_i}; q_i$ . As $e$ and $d$ are coprime the $p_i$ and the $q_i$ are distinct.

You have $x = (a - ek_1) = (a - k_1\prod p_i^{a_i})= (b - ek_2) = (b - k_2\prod q_i^{q_i})$

So $k_1\prod p_i^{a_i} = k_2\prod q_i^{q_i}$

So $\prod q_i^{b_i}|k_1$ and $\prod p_i^{a_i}|k_2$.

So $x = (a - k_3*\prod q_i^{b_i} \prod p_i^{a_i}) = (a- k_3de)$

fleablood
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