Find integers $t,z$ satisfying $(a+b)z+abt=1$

Question

Let be $a,b,u,v$ integers and $au+bv=1$. I have to find integers $t,z$ satisfying $(a+b)z+abt=1$.

I think I can solve it by congruence. What do you think?

I realised $$bz\equiv 1 \quad (\text{mod}\, a)\Longrightarrow z\equiv v \quad (\text{mod}\, a)$$

$$az\equiv 1 \quad (\text{mod}\, b)\Longrightarrow z\equiv u \quad (\text{mod}\, b)$$

Any ideas? Thank you.

The same question was asked with slightly different terminology in If $(a,b)=1$ then prove $(a+b, ab)=1$.. Several of the answers there start by rephrasing the question into your formulation. — hmakholm left over Monica, Mar 08 '14 at 12:35
But I want to know how to obtain $z$ and $t$. Not to shoe that they exist. — MaríaCC, Mar 08 '14 at 12:41
x @MaríaCC: At least two of the answers to the other question (by myself and Robjohn) explicitly derive formulas for the missing coefficients. — hmakholm left over Monica, Mar 08 '14 at 12:45

score 0 · Accepted Answer · edited Apr 13 '17 at 12:21

The first equation you wrote $1=au+bv$ is the Bézout's Identity and you can find $x$ and $y$ with the Eucledian Algorithm. More in general Bézout's Identity asserts that if there exists 2 integer numbers $m,n$ such that $GCD(m,n)=1$, then you can find 2 integer numbers $z,t$ such that $1=mz+nt$ with the Eucledian Algorithm. In this case $m=a+b, n=ab$. You can apply the Bézout's Identity because $GCD(a+b,ab)=1$...why? because $GCD(a,b)=1$, as explained here.

Find integers $t,z$ satisfying $(a+b)z+abt=1$

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