How can I solve a system of 2 congruences?

Question

I have this system of congruences $ p \equiv 3 \pmod 4$ and $ p \equiv 2 \pmod 3$ and the solution written in the book is $ p \equiv 11 \pmod {12 }$ but I do not know how? Could anyone explain this for me please?

I feel like I can not apply Chinese remainder theorem as it requires at least 3 congruences ..... am I correct?

What don't you understand? One of your tags refers to the Chinese Remainder Theorem...are you familiar with that theorem? — lulu, Jun 12 '19 at 17:58
No, it doesn't require three congruences. Just proceed as usual. — saulspatz, Jun 12 '19 at 17:59
Hint $\ 3,4\mid p+1\iff 3\cdot 4\mid p+1,$ by CCRT $\ \ $ — Bill Dubuque, Jun 12 '19 at 19:07

score 3 · Accepted Answer · answered Jun 12 '19 at 18:44

3

You could apply the Chinese remainder theorem. For this particular system, the solution is simple:

$p\equiv-1\pmod4$ and $p\equiv-1\pmod3$ means $4|p+1$ and $3|p+1$, so $12|p+1$;

i.e., $p\equiv-1\pmod{12}$.

[Note: $-1\equiv11\pmod{12}$.]

answered Jun 12 '19 at 18:44

J. W. Tanner

60,406

Very nice I did not think it will be very easy like this by CRT ..... Thanks a lot – Emptymind Jun 12 '19 at 23:38
Solving $p\equiv a \pmod n$ and $p\equiv a \pmod m$ is particularly easy; $p\equiv a \pmod n$ and $p\equiv b \pmod m$ requires a bit more effort – J. W. Tanner Jun 13 '19 at 00:49
1

Right. This is the constant case of CRT. – Jun 13 '19 at 21:50

score 2 · Answer 2 · answered Jun 12 '19 at 18:04

2

I like to use Bezout coefficients, like such: $-1\cdot3+1\cdot4=1$.

Then we write $x=3\cdot(-3)+2\cdot 4=-1\cong {11}\pmod{12}$.

answered Jun 12 '19 at 18:04

I can see in the first step that you are writing the gcd of 4 and 3 as a linear combination of 4 and 3 .... what about the second step of finding x? – Emptymind Jun 12 '19 at 18:08
Take that same equation and multiply the terms by the residues in the "opposite order". That is, I put the residue mod $4$ with the term with the $3$; and the residue mod $3$ with the term with the $4$. Do you follow? – Jun 12 '19 at 18:14
See Bernard's answer for a good visual. – Jun 13 '19 at 22:26

score 2 · Answer 3 · answered Jun 12 '19 at 18:10

For a problem like this, simple trial and error suffices.

For a general method: Start with $p\equiv 3\pmod 4$. That tells us that $p=3+4k$. Now we move to the second congruence, contingent on the first. $$p\equiv 2\pmod 3\iff 3+4k\equiv 2\pmod 3\iff k\equiv 2 \pmod 3$$ $$\iff k=2+3l\iff p = 11+12l\iff p\equiv 11 \pmod {12}$$

as desired.

score 2 · Answer 4 · answered Jun 12 '19 at 18:40

2

One starts from Bézout's identity: $\;4u+3v=1\quad(u,v\in\mathbf Z)$, say $4-3=1$. By the Chinese remainder theorem, $$\begin{cases} p\equiv \color{red}{\alpha\mod 4}\\ p\equiv\color{blue}{\beta\mod 3} \end{cases}\iff p\equiv \color{blue}{\beta}\,\color{red}{4}u+\color{red}{\alpha}\,\color{blue}{3}v\mod 3\cdot4 $$

answered Jun 12 '19 at 18:40

Bernard

175,478

I had a mind to do this (with the colors). But didn't follow through. – Jun 13 '19 at 21:48
1

Great minds think together ;o). Joking aside, I think it's a good way to insist on which value goes with which modulus in the solution formula. – Bernard Jun 13 '19 at 21:54
It's perfect for it. – Jun 13 '19 at 21:55
Thank you for your kind appreciation! – Bernard Jun 13 '19 at 21:57

score 1 · Answer 5 · answered Jun 12 '19 at 18:08

1

You can write $$p=3+4m,p=2+3n$$ so $$1=3n-4m$$ solving this Diophantine equation we get $$m=2+3C,n=4+4C$$ so $$p=11+12C$$

answered Jun 12 '19 at 18:08

Dr. Sonnhard Graubner

95,283

How can I solve a system of 2 congruences?

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