Chinese Remainder theorem for 2 congruences

[fishturtle1]

Homework Statement

Let $a, b, m, n$ be integers with $\gcd(m,n) = 1$. Let $$c \equiv (b-a)\cdot m^{-1} (\operatorname{mod} n)$$
Prove that $x = a + cn$ is a solution to $x \equiv a (\operatorname{mod} m)$ and $x \equiv b (\operatorname{mod} n)$, (2.24).

and that every solution to (2.24) has the form $x = a + cn + ymn$ for some $y \epsilon \mathbb{Z}$.

Homework Equations

The Attempt at a Solution

Since $\gcd(m,n) = 1$ I tried following CRT..

$x \equiv a (\operatorname{mod} m)$
$x = a + mk$, for some integer $k$.

Substituting into second congruence,
$a + mk \equiv b (\operatorname{mod} n)$
$mk \equiv b - a (\operatorname{mod} n)$
$k \equiv (b-a)\cdot m^{-1} (\operatorname{mod} n)$
$k = (b-a) \cdot m^{-1} + nl$, for some integer $l$.
Substituting $k$ into $x$,
$x = a + mk$
$x = a + m((b-a)\cdot m^{-1} + nl)$

But we're given $c = (b-a)\cdot m^{-1}$
So $x = a + mc + mnl$ but they have $nc$ instead of $mc$ as the second term of $x$Also, I tried to check that this $x$ satisfied both congruences but i have a problem:

We have $x \equiv a + mc \equiv a + m(b-a)\cdot m^{-1} \equiv a + b - a \equiv b (\operatorname{mod} mn)$.

So this satisifes $x = b \equiv b (\operatorname{mod} n)$
But i don't see how $x = b \equiv a (\operatorname{mod} m)$ is true unless $a \equiv b (\operatorname{mod} m)$...

 

[I like Serena]

It's a typo. The solution should be x=a+cm.
Then the first equation is satisfied trivially, and c cancels neatly in the 2nd equation.

Alternatively the typo could be in the given equations, meaning m and n should be swapped.

 

[fishturtle1]

Ok, thank you.