Exercise 3.16

Proof. Let $m_1=7,m_2=9,m_3=5,m=m_1{m}_2{m}_3=315,n_1=m∕m_1=m_2{m}_3=45,n_2=m_1{m}_3=35,n_3=m_1{m}_2=63$.

If $r_1=13,s_1=-2$, then $r_1{m}_1+s_1{n}_1=13m_1-2m_2{m}_3=13\times 7-2\times 45=1$,

so $e_1=s_1{n}_1=-2\times 45=-90$ verifies

If $r_2=4,s_2=-1$, then $r_2{m}_2+s_2{n}_2=4\times 9-1\times 35=1$,

so $e_2=s_2{n}_2=-35$ verifies

If $r_3=-25,s_3=2$, then $r_3{m}_3+s_3{n}_3=-25\times 5+2\times 63=1$,

so $e_3=s_3{n}_3=2\times 63=126$ verifies

Let $x_0=e_1+4e_2+3e_3=148$ : then

If $x\in \mathbb{Z}$ is any solution of the system, then $7\mid x-x_0,9\mid x-x_0,5\mid x-x_0$, with $7\wedge 9=7\wedge 5=9\wedge 5=1$, so $m=315\mid x-x_0$ :

and all these integers are solutions of the system. □