Find integers $b, c $ so that $3b\equiv3c\pmod{12}$, but $b \not\equiv c\pmod{12}$.

Question

$$3b\equiv{3c}\quad(\text{mod }12),$$ but $$b\not\equiv{c}\quad(\text{mod }12).$$

I am not sure how to approach this problem, any help is appreciated. Thanks!

Answer 1

$3b\equiv 3c\pmod{12}$ means $3b=3c+12k$ for k integer, and $b\not\equiv c\pmod{12}$ is saying $b\ne c+12k$'. Solve these inequalities: $0\ne 0+12k'-4k$, i.e., $4k\ne 12k'$, which again means $k\ne 3k'$. In summary, we have $b=c+4k$ for any c, k, with k is not multiple of 3.

Answer 2

$3 \times 4 = 0 = 3 \times 8 \pmod{12}$ but $4 \neq 8 \pmod{12}$.

We abuse that $3|12$, or more generally that $3$ and $12$ are not coprime.

Answer 3

Approach it by doing it.

Figure out what $3*a\mod 12$ is for $a=0...11$ and see if any of them are the same for different $a $s.

Doesn't take long to notice $3*0\equiv 3*4\mod 12$.

In fact for $a=0...11$ the $3a\equiv :$ $0,3,6,9,0,3,6,9,0,3,6,9$

So $0\equiv 3*0\equiv 3*4\equiv 3*8 \mod 12$

And $3\equiv 3*1\equiv 3*5\equiv 2*9\mod 12$

$6\equiv 3*2\equiv 3*6\equiv ... etc.$ Etc, are lots of counter examples.