How to solve system of congruence with common divisor?

Question

How to solve these (4) systems of congruences?

$$ \begin{cases} x=3,5\pmod{8}\\ x=5,7\pmod{12} \end{cases} $$ I was thinking about using CRT but $\gcd\left(8,12\right)>1$.

What can I do?

Clarification:

This notation represents 4 different systems. One of them for example is

$$ \begin{cases} x=3\pmod{8}\\ x=7\pmod{12} \end{cases} $$

@InterstellarProbe: Likely $x$ congruent to $3$ or to $5$ mod $8$, etc. So there are four separate problems tersely expressed. — hardmath, Jul 16 '18 at 13:35

score 2 · Accepted Answer · answered Jul 16 '18 at 13:50

2

As chí trunch châo mentioned. If you work $\textrm{mod}\ 24$, you should see that being $3\ \textrm{mod}\ 8$ gives $(3,11,19)\ \textrm{mod}\ 24$ (excuse my notation) and $5\ \textrm{mod}\ 8$ gives $(5,13,21)\ \textrm{mod}\ 24$. Moreover, $5\ \textrm{mod}\ 12$ gives $(5,17)\ \textrm{mod}\ 24$ and $7\ \textrm{mod}\ 12$ gives $(7,19)\ \textrm{mod}\ 24$. Looking at these possibilities and its intersection, it is not hard to see that the only two possibilities remaining are $5\ \textrm{mod}\ 24$ and $19\ \textrm{mod}\ 24$ which are solutions to the combinations $(5,5)$ and $(3,7)$, respectively.

answered Jul 16 '18 at 13:50

Stan Tendijck

2,432

Would this method work also for coprimes? – Jon Jul 16 '18 at 13:58
In general, you use the least common multiple (LCM). If they are coprime, you can even conclude that there always is a solution (Chinese Remainder Theorem) – Stan Tendijck Jul 16 '18 at 14:07
So, yes, it definitely works for coprimes – Stan Tendijck Jul 16 '18 at 14:07
Seems to be much faster looking for solutions this way than using CRT – Jon Jul 16 '18 at 14:08
Actually, the method uses the CRT – Stan Tendijck Jul 16 '18 at 14:19
@Jon We can simplify matters by notiing the residues are all $\pm 5,$ see my answer. – Bill Dubuque Feb 24 '19 at 02:28

score 1 · Answer 2 · answered Jul 16 '18 at 16:22

As observed, the result, if any, will be expressed to modulus of the lowest common multiple of the two given modulus values - here, $\text{lcm}(8,12)=24$.

For a given combination we can examine the stated equivalence in the modulus of the $\gcd$ to assess if it is feasible. Here $\gcd(8,12)=4$.

So for example: $$\left \{\begin{align} x&\equiv 5\bmod{8} & \implies x &\equiv 1\bmod 4\\ x&\equiv 7\bmod{12} & \implies x &\equiv 3\bmod 4\\ \end{align} \right . $$ and thus no solutions.

If this check shows that solutions are possible, as here: $$\left \{\begin{align} x&\equiv 3\bmod{8} & \implies x &\equiv 3\bmod 4\\ x&\equiv 7\bmod{12} & \implies x &\equiv 3\bmod 4\\ \end{align} \right . $$ we could proceed (for cases not as amenable to simple examination) by noting that the first equivalence means that $x= 8k+3$ and thus
$\begin{align} 8k+3 &\equiv 7\bmod 12\\ 8k &\equiv 4\bmod 12 \\ 2k &\equiv 1\bmod 3 &\text{(divide through all by gcd)}\\ k &\equiv 2\bmod 3 \\ \therefore x &\equiv 8\cdot 2 +3 \equiv 19 \bmod 24 \end{align}$

You missed the $\pm 5$ symmetry - see my answer. – Bill Dubuque Feb 24 '19 at 02:30 — Bill Dubuque, Feb 24 '19 at 02:30

score 0 · Answer 3 · answered Feb 24 '19 at 02:27

0

Simpler: it's $\ x\equiv \pm 5\bmod 8\ \&\ 12.$ We can't have $\,x\equiv -5,5\,$ or $\,5,-5\bmod 8,12$ else $\,x\equiv -1,1$ or $1,-1\bmod 4.\,$ So $\,x\equiv 5,5\,$ or $\,-5,-5\bmod 8,12\iff x\equiv \pm5\bmod 24,\,$ by $24={\rm lcm}(8,12)$

answered Feb 24 '19 at 02:27

Bill Dubuque

282,220

The final LCM inference is equivalent to CCRT = Constant case of CRT. – Bill Dubuque Feb 24 '19 at 02:32

How to solve system of congruence with common divisor?

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