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I have the following system of linear congruences: \begin{align} x&\equiv -2 \pmod {6}\\ x&\equiv 2 \pmod {10}\\ x&\equiv 7 \pmod {15} \end{align}

From the first equation we have: $ x= 6 k-2, k\in \mathbb{Z}$. Then we insert it into 2nd equation:

$ 6 k -2 \equiv 2 (\mod 10) $ i.e. $ 6 k \equiv 4 (\mod 10) $

from which we obtain

$ 3 k \equiv 2 (\mod 5) $

Now at this point, is it allowed to multiply the last congruence by 2 to get:

$ k \equiv -1 (\mod 5) $

and conclude $k = 5 l -1, l\in \mathbb{Z}$ and $ x = 30 n -8 , n\in \mathbb{Z}$ ?

I wonder if this is correct, is the obtained congruence equivalent to the first one? Thanks in advance.

Bill Dubuque
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user121
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  • What you've done is correct. As to your last question, $x\equiv-8\pmod30$ is consistent with $x\equiv -2\pmod6$, but not equivalent to it. I'm not sure you're asking the question you mean to ask. – saulspatz Apr 10 '21 at 12:21
  • @saulspatz If $(a,b)=d$, then $a \equiv b(\mod m) $ implies that $\frac{a}{d} \equiv \frac{b}{d}(\mod \frac{m}{d})$. But when we multiply by $d$, we get $ a \equiv b (\mod \frac{m}{d} )$. So, are solutions of $ 6 k \equiv 4 (mod 10) $ equivalent to solutions of $ k \equiv -1 (\mod 5) $? The 1st one implies the 2nd one, but does $ k \equiv -1 (\mod 5) $ imply $ 6 k \equiv 4 (mod 10) $ ? Or we don't need equivalences at all when solving such tasks? – user121 Apr 10 '21 at 12:41
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    We don't need equivalences, but implications (logical consequences). – saulspatz Apr 10 '21 at 12:58
  • Thanks a lot for explanation. – user121 Apr 10 '21 at 13:01
  • $k\equiv -1\pmod {5}\implies k\equiv 4,9\pmod {10}\implies 6k\equiv 24,54\pmod {10}$ but this doesn't technically cover all numbers that have remainder 4 mod 10 . – Roddy MacPhee Apr 10 '21 at 14:48
  • @RoddyMacPhee Does it mean that we generally cannot cover all solutions of systems if moduli are not relatively prime pairwise? (otherwise, I think that Chinese Remainder Theorem guarantees existence and uniqueness of solution) – user121 Apr 10 '21 at 14:58
  • more properly is 24 mod 30 . that's the piont. then you subtract the 2. – Roddy MacPhee Apr 10 '21 at 17:04

2 Answers2

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One must indeed be very careful to correctly keep track of implications in derivations like this. The key idea of this form of CRT is that we can solve a system of congruences by successively replacing a pair of congruences by an equivalent congruence, till we reach a single congruence. Thus it is essential to be sure that the replacement for each pair remains equivalent at each step.

So, e.g. in your case we seek to replace your first two congruences by the equivalent one below

$$\begin{align} x&\equiv -2\!\!\pmod{6}\\ x&\equiv\ \ \ 2\!\pmod{10}\end{align}\iff x\equiv -8\!\!\pmod{\!30}$$

There are a couple ways to prove that equivalence. First we could use all unidirectional inferences $(\Rightarrow)$ in our intermediate steps to obtain a proof of the direction $(\Rightarrow),\,$ then we could prove the direction $(\Leftarrow)$ by simply checking that $\,x\equiv 8\pmod{30}\,$ is actually a root of the two LHS congruences.

However this is actually not needed because every step of this (standard) method of CRT solution of a pair of congruences is actually an equivalence. Below I show the intermediate steps in detail.

$$\begin{align} &\ x\equiv -2\!\!\!\pmod{\!6},\ \ x\equiv 2\!\!\!\pmod{\!10}\\[.2em] \iff\ &\exists k\!:\, x = -2+6k,\ \ x\equiv 2\!\!\!\pmod{\!10}\\[.2em] \iff\ &\exists k\!:\, x = -2+6k,\ {-}2+6k\equiv 2\!\!\!\pmod{\!10}\\[.2em] \iff\ &\exists k\!:\, x = -2+6k,\ \ k\equiv -1\!\!\!\pmod{\!5}\\[.2em] \iff\ &\exists k\!:\, x = -2+6k,\ \exists n\!:\, k = -1+5n\\[.2em] \iff\ &\exists n\!:\, x = -2+6(-1+5n) = -8+30n\\ \iff\ &\, x\equiv -8\!\!\!\pmod{\!30} \end{align}$$

Remark $ $ For more on the ramifications of insufficiency of unidirectional inferences here (esp. its manifestation as extraneous roots) see here and its links.

Bill Dubuque
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Another approach:

$\begin{cases} x\equiv -2 \bmod 6\\x=2 \bmod 10\end{cases}$

$\rightarrow 6t_1-2=10t_2 +2\rightarrow 6t_1-10t_2=4 \rightarrow t_1=-10 s-1, t_2=6s -1$

Which gives:

$t_1=-10 s-1\rightarrow x=6(-10s-1)-2=-60 s -8$

$t_2=6s-1 \rightarrow x=10(6s-1)+2=60s -8$

$\begin{cases} x\equiv 7 \bmod 15\\x=2 \bmod 10\end{cases}$

$10t_3+2=15t_4+7$

which gives:

$t_3=15s_1-1\rightarrow x=10(15 s_1-1)+2=150 s_1-8$

$t_4=-10 s_1-1\rightarrow x=15(-10s_1-1)+7=-150 s_1-8$

$ lcm(6, 10, 15)=30|60, 150$, therefore:

$$x\equiv -8 \bmod 30\rightarrow x=30n -8$$

sirous
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