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Solve the system of congruence relations:

$$5x+17y\equiv 1 \pmod{24}$$

$$6x+10y\equiv 2\pmod{24}$$

Can someone help me? I usually use Chinese theorem, but here that is not possible.

Bill Dubuque
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math_
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  • No, I can't multiply first equation by 6 for example because 6 and 24 are not relatively prime. – math_ Jan 13 '21 at 17:30
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    You can solve the first congruence for either one of the variables (since both $5$ and $17$ are relatively prime to $24$). Then plugging that solution into the second congruence will give you necessary conditions for that variable, which can then be further investigated. – Greg Martin Jan 13 '21 at 17:32
  • You have, for the second, if you want, $3x+5y\equiv 1 \bmod 12$ (divide $6x+10y=2+24m$ by $2$). – Mark Bennet Jan 13 '21 at 17:36
  • @MarkBennet But what to do after I divide? – math_ Jan 13 '21 at 17:40
  • Reopened because the proposed dupe doesn't work here due to the singular determinant. – Bill Dubuque Jan 14 '21 at 13:09

1 Answers1

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Subtract both equations to make $x$ appear.

$\begin{cases}5x+17y\equiv 1\pmod{24}\\6x+10y\equiv 2\pmod{24}\end{cases}\iff \begin{cases}5x+17y\equiv 1\pmod{24}\\x\equiv 7y+1\pmod{24}\end{cases}$

We can now substitute in the first equation to solve for $y$.

$5x+17y\equiv 35y+5+17y\equiv 4y+5\equiv 1\pmod{24}\iff 4y\equiv 20\pmod{24}$

Now all coefficients are divisible by $4$ including the modulo, therefore we can simplify it to $$y\equiv 5\pmod 6$$

Note: do not make the mistake to go $\, y\equiv 5\pmod{24}\, $, this is not true, you cannot divide by $4$ because $\gcd(24,4)\neq 1$, but you can reduce the equation to a lower modulo when this one is also divisible.

We continue by setting $\, y=6n+5\, $ for some integer $n$.

$\begin{cases}y\equiv 6n+5\pmod{24}\\x\equiv 7y+1\pmod{24}\end{cases}\iff \begin{cases}y\equiv 6n+5\pmod{24}\\x\equiv 42n+36\pmod{24}\end{cases}\iff \begin{cases}y\equiv 6n+5\pmod{24}\\x\equiv 18n+12\pmod{24}\end{cases}$

You can verify the the solutions are cycling for $n\ge4$, you get for $n=0,1,2,3$

$$\{(12,5),(6,11),(0,17),(18,23)\}$$

zwim
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